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Finding value of $c$ such that the range of the rational function $f(x) = frac{x^2 + x + c}{x^2 + 2x + c}$...
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This question already has an answer here:
Finding the value of $c$
4 answers
I am comfortable when I am asked to calculate the range of a rational function, but how do we do the reverse? I came across this problem.
If $$f(x)= frac{x^2 + x + c}{x^2 + 2x + c}$$ then find the value of $c$ for which the range of $f(x)$ does not contain $[-1, -frac{1}{3}]$.
functions
edited Jan 5 at 16:16
N. F. Taussig
44k93355
asked Jan 5 at 15:58
Swap NayakSwap Nayak
22
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marked as duplicate by Saad, Abcd, A. Pongrácz, Holo, amWhy Jan 7 at 15:24
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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show 1 more comment
$begingroup$
This question already has an answer here:
Finding the value of $c$
4 answers
I am comfortable when I am asked to calculate the range of a rational function, but how do we do the reverse? I came across this problem.
If $$f(x)= frac{x^2 + x + c}{x^2 + 2x + c}$$ then find the value of $c$ for which the range of $f(x)$ does not contain $[-1, -frac{1}{3}]$.
functions
edited Jan 5 at 16:16
N. F. Taussig
44k93355
asked Jan 5 at 15:58
Swap NayakSwap Nayak
22
$endgroup$
marked as duplicate by Saad, Abcd, A. Pongrácz, Holo, amWhy Jan 7 at 15:24
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Find the range of this function as you usually would. You should get an answer in terms of $c$. Then match this to the requirement to fix $c$.
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– John Doe
Jan 5 at 16:08
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See math.stackexchange.com/questions/2759764/finding-the-value-of-c
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– lab bhattacharjee
Jan 5 at 16:14
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and math.stackexchange.com/questions/1414298/…
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– lab bhattacharjee
Jan 5 at 16:15
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I let the function equal to the and got a quadratic in x. I used the fact that x is real and thus set the discriminant greater than or equal to zero. I then got a quadratic inequality in y and c. This inequality should be broken when y is between -1 and -1/3. So I graphed the new function of y and checked where the inequality was broken and calculated the corresponding value of c.
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– Swap Nayak
Jan 5 at 16:32
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But it gives the wrong answer. I guess it is because there can be multiple graphs where the inequality is broken and so multiple values of c.
$endgroup$
– Swap Nayak
Jan 5 at 16:35
|
show 1 more comment
$begingroup$
This question already has an answer here:
Finding the value of $c$
4 answers
I am comfortable when I am asked to calculate the range of a rational function, but how do we do the reverse? I came across this problem.
If $$f(x)= frac{x^2 + x + c}{x^2 + 2x + c}$$ then find the value of $c$ for which the range of $f(x)$ does not contain $[-1, -frac{1}{3}]$.
functions
edited Jan 5 at 16:16
N. F. Taussig
44k93355
asked Jan 5 at 15:58
Swap NayakSwap Nayak
22
$endgroup$
This question already has an answer here:
Finding the value of $c$
4 answers
I am comfortable when I am asked to calculate the range of a rational function, but how do we do the reverse? I came across this problem.
If $$f(x)= frac{x^2 + x + c}{x^2 + 2x + c}$$ then find the value of $c$ for which the range of $f(x)$ does not contain $[-1, -frac{1}{3}]$.
This question already has an answer here:
Finding the value of $c$
4 answers
functions
functions
edited Jan 5 at 16:16
N. F. Taussig
44k93355
asked Jan 5 at 15:58
Swap NayakSwap Nayak
22
edited Jan 5 at 16:16
N. F. Taussig
44k93355
asked Jan 5 at 15:58
Swap NayakSwap Nayak
22
edited Jan 5 at 16:16
N. F. Taussig
44k93355
edited Jan 5 at 16:16
N. F. Taussig
44k93355
edited Jan 5 at 16:16
N. F. Taussig
44k93355
44k93355
asked Jan 5 at 15:58
Swap NayakSwap Nayak
22
asked Jan 5 at 15:58
Swap NayakSwap Nayak
22
asked Jan 5 at 15:58
Swap NayakSwap Nayak
22
22
marked as duplicate by Saad, Abcd, A. Pongrácz, Holo, amWhy Jan 7 at 15:24
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Saad, Abcd, A. Pongrácz, Holo, amWhy Jan 7 at 15:24
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Find the range of this function as you usually would. You should get an answer in terms of $c$. Then match this to the requirement to fix $c$.
$endgroup$
– John Doe
Jan 5 at 16:08
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See math.stackexchange.com/questions/2759764/finding-the-value-of-c
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– lab bhattacharjee
Jan 5 at 16:14
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and math.stackexchange.com/questions/1414298/…
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– lab bhattacharjee
Jan 5 at 16:15
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I let the function equal to the and got a quadratic in x. I used the fact that x is real and thus set the discriminant greater than or equal to zero. I then got a quadratic inequality in y and c. This inequality should be broken when y is between -1 and -1/3. So I graphed the new function of y and checked where the inequality was broken and calculated the corresponding value of c.
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– Swap Nayak
Jan 5 at 16:32
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But it gives the wrong answer. I guess it is because there can be multiple graphs where the inequality is broken and so multiple values of c.
$endgroup$
– Swap Nayak
Jan 5 at 16:35
|
show 1 more comment
$begingroup$
Find the range of this function as you usually would. You should get an answer in terms of $c$. Then match this to the requirement to fix $c$.
$endgroup$
– John Doe
Jan 5 at 16:08
$begingroup$
See math.stackexchange.com/questions/2759764/finding-the-value-of-c
$endgroup$
– lab bhattacharjee
Jan 5 at 16:14
$begingroup$
and math.stackexchange.com/questions/1414298/…
$endgroup$
– lab bhattacharjee
Jan 5 at 16:15
$begingroup$
I let the function equal to the and got a quadratic in x. I used the fact that x is real and thus set the discriminant greater than or equal to zero. I then got a quadratic inequality in y and c. This inequality should be broken when y is between -1 and -1/3. So I graphed the new function of y and checked where the inequality was broken and calculated the corresponding value of c.
$endgroup$
– Swap Nayak
Jan 5 at 16:32
$begingroup$
But it gives the wrong answer. I guess it is because there can be multiple graphs where the inequality is broken and so multiple values of c.
$endgroup$
– Swap Nayak
Jan 5 at 16:35
$begingroup$
Find the range of this function as you usually would. You should get an answer in terms of $c$. Then match this to the requirement to fix $c$.
$endgroup$
– John Doe
Jan 5 at 16:08
$begingroup$
Find the range of this function as you usually would. You should get an answer in terms of $c$. Then match this to the requirement to fix $c$.
$endgroup$
– John Doe
Jan 5 at 16:08
$begingroup$
See math.stackexchange.com/questions/2759764/finding-the-value-of-c
$endgroup$
– lab bhattacharjee
Jan 5 at 16:14
$begingroup$
See math.stackexchange.com/questions/2759764/finding-the-value-of-c
$endgroup$
– lab bhattacharjee
Jan 5 at 16:14
$begingroup$
and math.stackexchange.com/questions/1414298/…
$endgroup$
– lab bhattacharjee
Jan 5 at 16:15
$begingroup$
and math.stackexchange.com/questions/1414298/…
$endgroup$
– lab bhattacharjee
Jan 5 at 16:15
$begingroup$
I let the function equal to the and got a quadratic in x. I used the fact that x is real and thus set the discriminant greater than or equal to zero. I then got a quadratic inequality in y and c. This inequality should be broken when y is between -1 and -1/3. So I graphed the new function of y and checked where the inequality was broken and calculated the corresponding value of c.
$endgroup$
– Swap Nayak
Jan 5 at 16:32
$begingroup$
I let the function equal to the and got a quadratic in x. I used the fact that x is real and thus set the discriminant greater than or equal to zero. I then got a quadratic inequality in y and c. This inequality should be broken when y is between -1 and -1/3. So I graphed the new function of y and checked where the inequality was broken and calculated the corresponding value of c.
$endgroup$
– Swap Nayak
Jan 5 at 16:32
$begingroup$
But it gives the wrong answer. I guess it is because there can be multiple graphs where the inequality is broken and so multiple values of c.
$endgroup$
– Swap Nayak
Jan 5 at 16:35
$begingroup$
But it gives the wrong answer. I guess it is because there can be multiple graphs where the inequality is broken and so multiple values of c.
$endgroup$
– Swap Nayak
Jan 5 at 16:35
|
show 1 more comment
3 Answers
3
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oldest
votes
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We should probably proceed by cases, as $f$ will have a vertical asymptote and a hole for $c=0,$ a single vertical asymptote and no holes for $c=1,$ neither vertical asymptotes nor holes when $c>1,$ and otherwise has two vertical asymptotes and no holes. In all cases, $f$ has $y=1$ as its horizontal asymptote, meaning that $lim_{|x|toinfty}f(x)=1.$
In the $c>1$ case, differentiating yields $$f'(x)=frac{x^2-c}{(x^2+2x+c)^2},$$ which is negative in the interval $left(-sqrt{c},sqrt{c}right),$ zero at $x=pmsqrt{c},$ and positive otherwise. Consequently, $f$ achieves a global maximum at $x=-sqrt{c}$ and a global minimum at $x=sqrt{c}.$ The minimum is the one we care about, though, since we'll require it to be greater than $-1.$ So, let's consider $$-1<fleft(sqrt cright)=frac{c+sqrt c+c}{c+2sqrt c+c}=frac{2c+sqrt c}{2c+2sqrt c}=frac{2sqrt c+1}{2sqrt c+2}.$$ But this is trivially true, since $c>1$ implies that both $2sqrt c+1$ and $2sqrt c+2$ are positive, so that $f(c)>0>-1,$ as desired.
I leave the $c=0$ and $c=1$ cases to you.
If $c<1$ and $cne 0,$ we know that $x^2+2x+c=(x-j)(x-k)$ for some distinct real numbers $j$ and $k.$ We also know that $x^2+x+c$ takes on a non-zero value at each of $j,k.$ Without loss of generality, suppose that $j<k.$ Now, at $x=j,$ the denominator changes signs, while the numerator does not, meaning that we have either $$lim_{xnearrow j}f(x)=-inftytext{ and }lim_{xsearrow j}f(x)=+inftytag{1}$$ or $$lim_{xnearrow j}f(x)=+inftytext{ and }lim_{xsearrow j}f(x)=-infty.tag{2}$$ Likewise, we have either $$lim_{xnearrow k}f(x)=-inftytext{ and }lim_{xsearrow k}f(x)=+inftytag{3}$$ or $$lim_{xnearrow k}f(x)=+inftytext{ and }lim_{xsearrow k}f(x)=-infty.tag{4}$$ If $(1)$ holds, then since $f$ is continuous on $(-infty,j)$ and since $lim_{xto-infty}f(x)=1$ and $lim_{xnearrow j}f(x)=-infty,$ then $(-infty,1)$ is a subset of the range of $f,$ and so $left[-1,-frac13right]$ is, too. If $(4)$ holds, then by continuity of $f$ on $(k,infty),$ we similarly have that $left[-1,-frac13right]$ is a subset of the range of $f.$ However, if $(2)$ and $(3)$ both hold, it isn't clear from what we've discussed so far whether $left[-1,-frac13right]$ is a subset of the range of $f$ or not, so let's consider further when these different possibilities may occur.
The only way for $(1)$ to hold is if the numerator is negative at $x=j,$ meaning that $j$ must lie between the two zeroes of $x^2+x+c.$ In other words, by the quadratic formula, we have the following (equivalent) inequalities: $$frac{-1-sqrt{1^2-4c}}2<frac{-2-sqrt{2^2-4c}}2<frac{-1+sqrt{1^2-4c}}2\-1-sqrt{1-4c}<-2-sqrt{4-4c}<-1+sqrt{1-4c}\-sqrt{1-4c}<-1-sqrt{4-4c}<sqrt{1-4c}\left(-1-sqrt{4-4c}right)^2<1-4c\1+2sqrt{4-4c}+4-4c<1-4c\2sqrt{4-4c}+4<0.$$ Since this is impossible, then $(1)$ cannot hold, and so $(2)$ must hold.
Similarly, $(4)$ only holds if the numerator is negative at $x=k,$ which happens if and only if $$frac{-1-sqrt{1^2-4c}}2<frac{-2+sqrt{2^2-4c}}2<frac{-1+sqrt{1^2-4c}}2\-1-sqrt{1-4c}<-2+sqrt{4-4c}<-1+sqrt{1-4c}\-sqrt{1-4c}<-1+sqrt{4-4c}<sqrt{1-4c}\left(-1+sqrt{4-4c}right)^2<1-4c\1-2sqrt{4-4c}+4-4c<1-4c\-2sqrt{4-4c}+4<0\2<sqrt{4-4c}\4<4-4c\c<0.$$ Thus, if $c<0,$ then $left[-1,-frac13right]$ is a subset of the range of $f.$
It remains only to consider $0<c<1,$ in which case $(2)$ and $(3)$ will hold, so that $f$ achieves a local maximum in the interval $(j,k).$ Observe that $$1-c<1\sqrt{1-c}<1\2sqrt{1-c}<2\sqrt{4-4c}<2\-2+sqrt{4-4c}<0\frac{-2+sqrt{4-4c}}2<0\k<0,$$ meaning that $f$ achieves its local maximum at $x=-sqrt{c},$ and a local minimum at $x=sqrt{c}$ by the first derivative test. Since $f$ is increasing and continuous on $(-infty,j),$ and since $lim_{xto-infty}f(x)=1,$ then $f$ is positive on $(-infty,j).$ Further, since $fbigl(sqrt cbigr)$ is positive by prior work, then since $f$'s minimum value in $(k,infty)$ occurs at $x=sqrt c,$ then we have that $f$ is positive on $(k,infty).$ Thus, we need only ensure that $fleft(-sqrt cright)<-frac13.$ Since $0<c<1,$ then $c<sqrt c,$ and so the following are equivalent: $$fleft(-sqrt cright)<-frac13\-3fleft(-sqrt cright)>1\-3cdotfrac{c-sqrt c+c}{c-2sqrt c+c}>1\-3cdotfrac{2c-sqrt c}{2c-2sqrt c}>1\-3left(2c-sqrt cright)<2c-2sqrt c\-6c+3sqrt c<2c-2sqrt c\5sqrt c<8c\frac58<sqrt c\frac{25}{64}<c.$$
Putting it all together, we find that the range of $f$ does not contain the interval in question if and only if $c>frac{25}{64}.$
answered Jan 5 at 19:09
Cameron BuieCameron Buie
85.1k771155
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but when $c=frac12>frac{25}{64}$, it does contain the given interval, see the graph.
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– farruhota
Jan 5 at 19:56
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@farruhota: Incorrect. It contains points of the interval (as elements), but does not contain the interval, itself (as a subset). See the graph, yourself, and you'll find that the range is a subset of $left(-infty,-sqrt2right]cup[0,+infty),$ which does not (for example) contain $-frac13.$
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– Cameron Buie
Jan 5 at 20:01
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Agreed +1, there is a subtle difference between "completely" and "partially". In that case I must set my condition as yours $f(-sqrt{c})<-frac13$ to get the same interval for $c$ as yours. Thank you.
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– farruhota
Jan 5 at 20:11
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Even I got the same answer c>25/64. But it's wrong putting c as 26/64 f(x) is lying in the interval.
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– Swap Nayak
Jan 8 at 5:09
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@Swap See the above discussion with farruhota, as well as the discussion with John Doe. The question (as posed) is ambiguously worded.
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– Cameron Buie
Jan 8 at 7:53
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Refer to the Desmos graph.
Find the local maximum:
$$begin{align}f'(x)&= left(frac{x^2 + x + c}{x^2 + 2x + c}right)'=\
&=frac{(2x+1)(x^2+2x+c)-(x^2+x+c)(2x+2)}{(x^2 + 2x + c)^2}=0 Rightarrow \
x^2&=c Rightarrow x=pm sqrt{c}end{align}$$
Note the local maximum occurs for $x=-sqrt{c}$, for which:
$$f(-sqrt{c})=frac{2sqrt{c}-1}{2sqrt{c}-2}<-1 Rightarrow cin left(frac9{16},1right).$$
For $cin [1,+infty)$:
$$f(x)=frac{x^2+x+c}{x^2+2x+c}=1-frac{x}{x^2+2x+c}>0.$$
Hence, when $c>frac9{16}$, the function's range will not contain $[-1,-frac13]$.
answered Jan 5 at 19:51
farruhotafarruhota
19.8k2738
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The inequality that you got should have been $$g(y,c)=4(1-c)y^2+2(4c-2)y+(1-4c)ge0$$Well, we want this to fail when $y$ is in this range. i.e. for $yin[-1,-1/3]$, $$4(1-c)y^2+2(4c-2)y+(1-4c)<0$$ This means if we plot this with $y$ on the $x$-axis and the value of $g(y)$ on the $y$-axis, then it will be a positive quadratic which dips below the $x$-axis for the range of $x$ values $[-1,-1/3]$. As we vary $c$, the places where this graph cuts the $x$ axis move (i.e. the roots of $g$). Note, when $c=0$, the graph is $$g(y,c=0)=4y^2-4y+1=(2y-1)^2$$which has one repeated root at $y=1/2$. Then as $c$ is increased, the minimum point of this quadratic moves leftward. The point at which $-1/3$ joins the range of values for which $g(-1/3,c)le0$ is when it first becomes a root. For what value of $c$ does this happen?
$$g(-1/3,c)=0impliescdotsimplies c=frac{25}{64}$$As we increase $c$ further, eventually $-1$ also joins this range. $$g(-1,c)=0impliescdotsimplies c=frac9{16}$$
For $c>frac9{16}$, $[-1,-1/3]$ is still not in the range of $f(x)$ - but $c=frac9{16}$ is the first value of $c$ for which this happens.
So the range of $c$ for which no points from the interval $[-1,-1/3]$ are included in the range of $f$ is given by $$bbox[5px,border:2px solid red]{cgefrac9{16}}$$
Meanwhile, the range of $c$ for which at least one point from the interval $[-1,-1/3]$ is missing from the range of $f$ is given by $$bbox[5px,border:2px solid red]{cgefrac{25}{64}}$$
answered Jan 5 at 17:35
John DoeJohn Doe
11.1k11238
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This seems to be trying to find the values of $c$ for which the range of $f$ is disjoint from the interval in question, rather than simply not containing the interval
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– Cameron Buie
Jan 5 at 19:04
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I assumed the question was asking for the values of $c$ for which the entire interval is not in the range, but I see what you mean - it could easily also mean the range of $c$ for which any point in this interval is not in the range. In this case, this cutoff would be at $c=frac{25}{64}$, which is the point at which the point $y=-1/3$ joins the range of values for which $g(y,c)le0$.
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– John Doe
Jan 7 at 14:29
add a comment |
3 Answers
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3 Answers
3
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oldest
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active
oldest
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active
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$begingroup$
We should probably proceed by cases, as $f$ will have a vertical asymptote and a hole for $c=0,$ a single vertical asymptote and no holes for $c=1,$ neither vertical asymptotes nor holes when $c>1,$ and otherwise has two vertical asymptotes and no holes. In all cases, $f$ has $y=1$ as its horizontal asymptote, meaning that $lim_{|x|toinfty}f(x)=1.$
In the $c>1$ case, differentiating yields $$f'(x)=frac{x^2-c}{(x^2+2x+c)^2},$$ which is negative in the interval $left(-sqrt{c},sqrt{c}right),$ zero at $x=pmsqrt{c},$ and positive otherwise. Consequently, $f$ achieves a global maximum at $x=-sqrt{c}$ and a global minimum at $x=sqrt{c}.$ The minimum is the one we care about, though, since we'll require it to be greater than $-1.$ So, let's consider $$-1<fleft(sqrt cright)=frac{c+sqrt c+c}{c+2sqrt c+c}=frac{2c+sqrt c}{2c+2sqrt c}=frac{2sqrt c+1}{2sqrt c+2}.$$ But this is trivially true, since $c>1$ implies that both $2sqrt c+1$ and $2sqrt c+2$ are positive, so that $f(c)>0>-1,$ as desired.
I leave the $c=0$ and $c=1$ cases to you.
If $c<1$ and $cne 0,$ we know that $x^2+2x+c=(x-j)(x-k)$ for some distinct real numbers $j$ and $k.$ We also know that $x^2+x+c$ takes on a non-zero value at each of $j,k.$ Without loss of generality, suppose that $j<k.$ Now, at $x=j,$ the denominator changes signs, while the numerator does not, meaning that we have either $$lim_{xnearrow j}f(x)=-inftytext{ and }lim_{xsearrow j}f(x)=+inftytag{1}$$ or $$lim_{xnearrow j}f(x)=+inftytext{ and }lim_{xsearrow j}f(x)=-infty.tag{2}$$ Likewise, we have either $$lim_{xnearrow k}f(x)=-inftytext{ and }lim_{xsearrow k}f(x)=+inftytag{3}$$ or $$lim_{xnearrow k}f(x)=+inftytext{ and }lim_{xsearrow k}f(x)=-infty.tag{4}$$ If $(1)$ holds, then since $f$ is continuous on $(-infty,j)$ and since $lim_{xto-infty}f(x)=1$ and $lim_{xnearrow j}f(x)=-infty,$ then $(-infty,1)$ is a subset of the range of $f,$ and so $left[-1,-frac13right]$ is, too. If $(4)$ holds, then by continuity of $f$ on $(k,infty),$ we similarly have that $left[-1,-frac13right]$ is a subset of the range of $f.$ However, if $(2)$ and $(3)$ both hold, it isn't clear from what we've discussed so far whether $left[-1,-frac13right]$ is a subset of the range of $f$ or not, so let's consider further when these different possibilities may occur.
The only way for $(1)$ to hold is if the numerator is negative at $x=j,$ meaning that $j$ must lie between the two zeroes of $x^2+x+c.$ In other words, by the quadratic formula, we have the following (equivalent) inequalities: $$frac{-1-sqrt{1^2-4c}}2<frac{-2-sqrt{2^2-4c}}2<frac{-1+sqrt{1^2-4c}}2\-1-sqrt{1-4c}<-2-sqrt{4-4c}<-1+sqrt{1-4c}\-sqrt{1-4c}<-1-sqrt{4-4c}<sqrt{1-4c}\left(-1-sqrt{4-4c}right)^2<1-4c\1+2sqrt{4-4c}+4-4c<1-4c\2sqrt{4-4c}+4<0.$$ Since this is impossible, then $(1)$ cannot hold, and so $(2)$ must hold.
Similarly, $(4)$ only holds if the numerator is negative at $x=k,$ which happens if and only if $$frac{-1-sqrt{1^2-4c}}2<frac{-2+sqrt{2^2-4c}}2<frac{-1+sqrt{1^2-4c}}2\-1-sqrt{1-4c}<-2+sqrt{4-4c}<-1+sqrt{1-4c}\-sqrt{1-4c}<-1+sqrt{4-4c}<sqrt{1-4c}\left(-1+sqrt{4-4c}right)^2<1-4c\1-2sqrt{4-4c}+4-4c<1-4c\-2sqrt{4-4c}+4<0\2<sqrt{4-4c}\4<4-4c\c<0.$$ Thus, if $c<0,$ then $left[-1,-frac13right]$ is a subset of the range of $f.$
It remains only to consider $0<c<1,$ in which case $(2)$ and $(3)$ will hold, so that $f$ achieves a local maximum in the interval $(j,k).$ Observe that $$1-c<1\sqrt{1-c}<1\2sqrt{1-c}<2\sqrt{4-4c}<2\-2+sqrt{4-4c}<0\frac{-2+sqrt{4-4c}}2<0\k<0,$$ meaning that $f$ achieves its local maximum at $x=-sqrt{c},$ and a local minimum at $x=sqrt{c}$ by the first derivative test. Since $f$ is increasing and continuous on $(-infty,j),$ and since $lim_{xto-infty}f(x)=1,$ then $f$ is positive on $(-infty,j).$ Further, since $fbigl(sqrt cbigr)$ is positive by prior work, then since $f$'s minimum value in $(k,infty)$ occurs at $x=sqrt c,$ then we have that $f$ is positive on $(k,infty).$ Thus, we need only ensure that $fleft(-sqrt cright)<-frac13.$ Since $0<c<1,$ then $c<sqrt c,$ and so the following are equivalent: $$fleft(-sqrt cright)<-frac13\-3fleft(-sqrt cright)>1\-3cdotfrac{c-sqrt c+c}{c-2sqrt c+c}>1\-3cdotfrac{2c-sqrt c}{2c-2sqrt c}>1\-3left(2c-sqrt cright)<2c-2sqrt c\-6c+3sqrt c<2c-2sqrt c\5sqrt c<8c\frac58<sqrt c\frac{25}{64}<c.$$
Putting it all together, we find that the range of $f$ does not contain the interval in question if and only if $c>frac{25}{64}.$
answered Jan 5 at 19:09
Cameron BuieCameron Buie
85.1k771155
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but when $c=frac12>frac{25}{64}$, it does contain the given interval, see the graph.
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– farruhota
Jan 5 at 19:56
1
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@farruhota: Incorrect. It contains points of the interval (as elements), but does not contain the interval, itself (as a subset). See the graph, yourself, and you'll find that the range is a subset of $left(-infty,-sqrt2right]cup[0,+infty),$ which does not (for example) contain $-frac13.$
$endgroup$
– Cameron Buie
Jan 5 at 20:01
$begingroup$
Agreed +1, there is a subtle difference between "completely" and "partially". In that case I must set my condition as yours $f(-sqrt{c})<-frac13$ to get the same interval for $c$ as yours. Thank you.
$endgroup$
– farruhota
Jan 5 at 20:11
$begingroup$
Even I got the same answer c>25/64. But it's wrong putting c as 26/64 f(x) is lying in the interval.
$endgroup$
– Swap Nayak
Jan 8 at 5:09
$begingroup$
@Swap See the above discussion with farruhota, as well as the discussion with John Doe. The question (as posed) is ambiguously worded.
$endgroup$
– Cameron Buie
Jan 8 at 7:53
add a comment |
$begingroup$
We should probably proceed by cases, as $f$ will have a vertical asymptote and a hole for $c=0,$ a single vertical asymptote and no holes for $c=1,$ neither vertical asymptotes nor holes when $c>1,$ and otherwise has two vertical asymptotes and no holes. In all cases, $f$ has $y=1$ as its horizontal asymptote, meaning that $lim_{|x|toinfty}f(x)=1.$
In the $c>1$ case, differentiating yields $$f'(x)=frac{x^2-c}{(x^2+2x+c)^2},$$ which is negative in the interval $left(-sqrt{c},sqrt{c}right),$ zero at $x=pmsqrt{c},$ and positive otherwise. Consequently, $f$ achieves a global maximum at $x=-sqrt{c}$ and a global minimum at $x=sqrt{c}.$ The minimum is the one we care about, though, since we'll require it to be greater than $-1.$ So, let's consider $$-1<fleft(sqrt cright)=frac{c+sqrt c+c}{c+2sqrt c+c}=frac{2c+sqrt c}{2c+2sqrt c}=frac{2sqrt c+1}{2sqrt c+2}.$$ But this is trivially true, since $c>1$ implies that both $2sqrt c+1$ and $2sqrt c+2$ are positive, so that $f(c)>0>-1,$ as desired.
I leave the $c=0$ and $c=1$ cases to you.
If $c<1$ and $cne 0,$ we know that $x^2+2x+c=(x-j)(x-k)$ for some distinct real numbers $j$ and $k.$ We also know that $x^2+x+c$ takes on a non-zero value at each of $j,k.$ Without loss of generality, suppose that $j<k.$ Now, at $x=j,$ the denominator changes signs, while the numerator does not, meaning that we have either $$lim_{xnearrow j}f(x)=-inftytext{ and }lim_{xsearrow j}f(x)=+inftytag{1}$$ or $$lim_{xnearrow j}f(x)=+inftytext{ and }lim_{xsearrow j}f(x)=-infty.tag{2}$$ Likewise, we have either $$lim_{xnearrow k}f(x)=-inftytext{ and }lim_{xsearrow k}f(x)=+inftytag{3}$$ or $$lim_{xnearrow k}f(x)=+inftytext{ and }lim_{xsearrow k}f(x)=-infty.tag{4}$$ If $(1)$ holds, then since $f$ is continuous on $(-infty,j)$ and since $lim_{xto-infty}f(x)=1$ and $lim_{xnearrow j}f(x)=-infty,$ then $(-infty,1)$ is a subset of the range of $f,$ and so $left[-1,-frac13right]$ is, too. If $(4)$ holds, then by continuity of $f$ on $(k,infty),$ we similarly have that $left[-1,-frac13right]$ is a subset of the range of $f.$ However, if $(2)$ and $(3)$ both hold, it isn't clear from what we've discussed so far whether $left[-1,-frac13right]$ is a subset of the range of $f$ or not, so let's consider further when these different possibilities may occur.
The only way for $(1)$ to hold is if the numerator is negative at $x=j,$ meaning that $j$ must lie between the two zeroes of $x^2+x+c.$ In other words, by the quadratic formula, we have the following (equivalent) inequalities: $$frac{-1-sqrt{1^2-4c}}2<frac{-2-sqrt{2^2-4c}}2<frac{-1+sqrt{1^2-4c}}2\-1-sqrt{1-4c}<-2-sqrt{4-4c}<-1+sqrt{1-4c}\-sqrt{1-4c}<-1-sqrt{4-4c}<sqrt{1-4c}\left(-1-sqrt{4-4c}right)^2<1-4c\1+2sqrt{4-4c}+4-4c<1-4c\2sqrt{4-4c}+4<0.$$ Since this is impossible, then $(1)$ cannot hold, and so $(2)$ must hold.
Similarly, $(4)$ only holds if the numerator is negative at $x=k,$ which happens if and only if $$frac{-1-sqrt{1^2-4c}}2<frac{-2+sqrt{2^2-4c}}2<frac{-1+sqrt{1^2-4c}}2\-1-sqrt{1-4c}<-2+sqrt{4-4c}<-1+sqrt{1-4c}\-sqrt{1-4c}<-1+sqrt{4-4c}<sqrt{1-4c}\left(-1+sqrt{4-4c}right)^2<1-4c\1-2sqrt{4-4c}+4-4c<1-4c\-2sqrt{4-4c}+4<0\2<sqrt{4-4c}\4<4-4c\c<0.$$ Thus, if $c<0,$ then $left[-1,-frac13right]$ is a subset of the range of $f.$
It remains only to consider $0<c<1,$ in which case $(2)$ and $(3)$ will hold, so that $f$ achieves a local maximum in the interval $(j,k).$ Observe that $$1-c<1\sqrt{1-c}<1\2sqrt{1-c}<2\sqrt{4-4c}<2\-2+sqrt{4-4c}<0\frac{-2+sqrt{4-4c}}2<0\k<0,$$ meaning that $f$ achieves its local maximum at $x=-sqrt{c},$ and a local minimum at $x=sqrt{c}$ by the first derivative test. Since $f$ is increasing and continuous on $(-infty,j),$ and since $lim_{xto-infty}f(x)=1,$ then $f$ is positive on $(-infty,j).$ Further, since $fbigl(sqrt cbigr)$ is positive by prior work, then since $f$'s minimum value in $(k,infty)$ occurs at $x=sqrt c,$ then we have that $f$ is positive on $(k,infty).$ Thus, we need only ensure that $fleft(-sqrt cright)<-frac13.$ Since $0<c<1,$ then $c<sqrt c,$ and so the following are equivalent: $$fleft(-sqrt cright)<-frac13\-3fleft(-sqrt cright)>1\-3cdotfrac{c-sqrt c+c}{c-2sqrt c+c}>1\-3cdotfrac{2c-sqrt c}{2c-2sqrt c}>1\-3left(2c-sqrt cright)<2c-2sqrt c\-6c+3sqrt c<2c-2sqrt c\5sqrt c<8c\frac58<sqrt c\frac{25}{64}<c.$$
Putting it all together, we find that the range of $f$ does not contain the interval in question if and only if $c>frac{25}{64}.$
answered Jan 5 at 19:09
Cameron BuieCameron Buie
85.1k771155
$endgroup$
$begingroup$
but when $c=frac12>frac{25}{64}$, it does contain the given interval, see the graph.
$endgroup$
– farruhota
Jan 5 at 19:56
1
$begingroup$
@farruhota: Incorrect. It contains points of the interval (as elements), but does not contain the interval, itself (as a subset). See the graph, yourself, and you'll find that the range is a subset of $left(-infty,-sqrt2right]cup[0,+infty),$ which does not (for example) contain $-frac13.$
$endgroup$
– Cameron Buie
Jan 5 at 20:01
$begingroup$
Agreed +1, there is a subtle difference between "completely" and "partially". In that case I must set my condition as yours $f(-sqrt{c})<-frac13$ to get the same interval for $c$ as yours. Thank you.
$endgroup$
– farruhota
Jan 5 at 20:11
$begingroup$
Even I got the same answer c>25/64. But it's wrong putting c as 26/64 f(x) is lying in the interval.
$endgroup$
– Swap Nayak
Jan 8 at 5:09
$begingroup$
@Swap See the above discussion with farruhota, as well as the discussion with John Doe. The question (as posed) is ambiguously worded.
$endgroup$
– Cameron Buie
Jan 8 at 7:53
add a comment |
$begingroup$
We should probably proceed by cases, as $f$ will have a vertical asymptote and a hole for $c=0,$ a single vertical asymptote and no holes for $c=1,$ neither vertical asymptotes nor holes when $c>1,$ and otherwise has two vertical asymptotes and no holes. In all cases, $f$ has $y=1$ as its horizontal asymptote, meaning that $lim_{|x|toinfty}f(x)=1.$
In the $c>1$ case, differentiating yields $$f'(x)=frac{x^2-c}{(x^2+2x+c)^2},$$ which is negative in the interval $left(-sqrt{c},sqrt{c}right),$ zero at $x=pmsqrt{c},$ and positive otherwise. Consequently, $f$ achieves a global maximum at $x=-sqrt{c}$ and a global minimum at $x=sqrt{c}.$ The minimum is the one we care about, though, since we'll require it to be greater than $-1.$ So, let's consider $$-1<fleft(sqrt cright)=frac{c+sqrt c+c}{c+2sqrt c+c}=frac{2c+sqrt c}{2c+2sqrt c}=frac{2sqrt c+1}{2sqrt c+2}.$$ But this is trivially true, since $c>1$ implies that both $2sqrt c+1$ and $2sqrt c+2$ are positive, so that $f(c)>0>-1,$ as desired.
I leave the $c=0$ and $c=1$ cases to you.
If $c<1$ and $cne 0,$ we know that $x^2+2x+c=(x-j)(x-k)$ for some distinct real numbers $j$ and $k.$ We also know that $x^2+x+c$ takes on a non-zero value at each of $j,k.$ Without loss of generality, suppose that $j<k.$ Now, at $x=j,$ the denominator changes signs, while the numerator does not, meaning that we have either $$lim_{xnearrow j}f(x)=-inftytext{ and }lim_{xsearrow j}f(x)=+inftytag{1}$$ or $$lim_{xnearrow j}f(x)=+inftytext{ and }lim_{xsearrow j}f(x)=-infty.tag{2}$$ Likewise, we have either $$lim_{xnearrow k}f(x)=-inftytext{ and }lim_{xsearrow k}f(x)=+inftytag{3}$$ or $$lim_{xnearrow k}f(x)=+inftytext{ and }lim_{xsearrow k}f(x)=-infty.tag{4}$$ If $(1)$ holds, then since $f$ is continuous on $(-infty,j)$ and since $lim_{xto-infty}f(x)=1$ and $lim_{xnearrow j}f(x)=-infty,$ then $(-infty,1)$ is a subset of the range of $f,$ and so $left[-1,-frac13right]$ is, too. If $(4)$ holds, then by continuity of $f$ on $(k,infty),$ we similarly have that $left[-1,-frac13right]$ is a subset of the range of $f.$ However, if $(2)$ and $(3)$ both hold, it isn't clear from what we've discussed so far whether $left[-1,-frac13right]$ is a subset of the range of $f$ or not, so let's consider further when these different possibilities may occur.
The only way for $(1)$ to hold is if the numerator is negative at $x=j,$ meaning that $j$ must lie between the two zeroes of $x^2+x+c.$ In other words, by the quadratic formula, we have the following (equivalent) inequalities: $$frac{-1-sqrt{1^2-4c}}2<frac{-2-sqrt{2^2-4c}}2<frac{-1+sqrt{1^2-4c}}2\-1-sqrt{1-4c}<-2-sqrt{4-4c}<-1+sqrt{1-4c}\-sqrt{1-4c}<-1-sqrt{4-4c}<sqrt{1-4c}\left(-1-sqrt{4-4c}right)^2<1-4c\1+2sqrt{4-4c}+4-4c<1-4c\2sqrt{4-4c}+4<0.$$ Since this is impossible, then $(1)$ cannot hold, and so $(2)$ must hold.
Similarly, $(4)$ only holds if the numerator is negative at $x=k,$ which happens if and only if $$frac{-1-sqrt{1^2-4c}}2<frac{-2+sqrt{2^2-4c}}2<frac{-1+sqrt{1^2-4c}}2\-1-sqrt{1-4c}<-2+sqrt{4-4c}<-1+sqrt{1-4c}\-sqrt{1-4c}<-1+sqrt{4-4c}<sqrt{1-4c}\left(-1+sqrt{4-4c}right)^2<1-4c\1-2sqrt{4-4c}+4-4c<1-4c\-2sqrt{4-4c}+4<0\2<sqrt{4-4c}\4<4-4c\c<0.$$ Thus, if $c<0,$ then $left[-1,-frac13right]$ is a subset of the range of $f.$
It remains only to consider $0<c<1,$ in which case $(2)$ and $(3)$ will hold, so that $f$ achieves a local maximum in the interval $(j,k).$ Observe that $$1-c<1\sqrt{1-c}<1\2sqrt{1-c}<2\sqrt{4-4c}<2\-2+sqrt{4-4c}<0\frac{-2+sqrt{4-4c}}2<0\k<0,$$ meaning that $f$ achieves its local maximum at $x=-sqrt{c},$ and a local minimum at $x=sqrt{c}$ by the first derivative test. Since $f$ is increasing and continuous on $(-infty,j),$ and since $lim_{xto-infty}f(x)=1,$ then $f$ is positive on $(-infty,j).$ Further, since $fbigl(sqrt cbigr)$ is positive by prior work, then since $f$'s minimum value in $(k,infty)$ occurs at $x=sqrt c,$ then we have that $f$ is positive on $(k,infty).$ Thus, we need only ensure that $fleft(-sqrt cright)<-frac13.$ Since $0<c<1,$ then $c<sqrt c,$ and so the following are equivalent: $$fleft(-sqrt cright)<-frac13\-3fleft(-sqrt cright)>1\-3cdotfrac{c-sqrt c+c}{c-2sqrt c+c}>1\-3cdotfrac{2c-sqrt c}{2c-2sqrt c}>1\-3left(2c-sqrt cright)<2c-2sqrt c\-6c+3sqrt c<2c-2sqrt c\5sqrt c<8c\frac58<sqrt c\frac{25}{64}<c.$$
Putting it all together, we find that the range of $f$ does not contain the interval in question if and only if $c>frac{25}{64}.$
answered Jan 5 at 19:09
Cameron BuieCameron Buie
85.1k771155
$endgroup$
We should probably proceed by cases, as $f$ will have a vertical asymptote and a hole for $c=0,$ a single vertical asymptote and no holes for $c=1,$ neither vertical asymptotes nor holes when $c>1,$ and otherwise has two vertical asymptotes and no holes. In all cases, $f$ has $y=1$ as its horizontal asymptote, meaning that $lim_{|x|toinfty}f(x)=1.$
In the $c>1$ case, differentiating yields $$f'(x)=frac{x^2-c}{(x^2+2x+c)^2},$$ which is negative in the interval $left(-sqrt{c},sqrt{c}right),$ zero at $x=pmsqrt{c},$ and positive otherwise. Consequently, $f$ achieves a global maximum at $x=-sqrt{c}$ and a global minimum at $x=sqrt{c}.$ The minimum is the one we care about, though, since we'll require it to be greater than $-1.$ So, let's consider $$-1<fleft(sqrt cright)=frac{c+sqrt c+c}{c+2sqrt c+c}=frac{2c+sqrt c}{2c+2sqrt c}=frac{2sqrt c+1}{2sqrt c+2}.$$ But this is trivially true, since $c>1$ implies that both $2sqrt c+1$ and $2sqrt c+2$ are positive, so that $f(c)>0>-1,$ as desired.
I leave the $c=0$ and $c=1$ cases to you.
If $c<1$ and $cne 0,$ we know that $x^2+2x+c=(x-j)(x-k)$ for some distinct real numbers $j$ and $k.$ We also know that $x^2+x+c$ takes on a non-zero value at each of $j,k.$ Without loss of generality, suppose that $j<k.$ Now, at $x=j,$ the denominator changes signs, while the numerator does not, meaning that we have either $$lim_{xnearrow j}f(x)=-inftytext{ and }lim_{xsearrow j}f(x)=+inftytag{1}$$ or $$lim_{xnearrow j}f(x)=+inftytext{ and }lim_{xsearrow j}f(x)=-infty.tag{2}$$ Likewise, we have either $$lim_{xnearrow k}f(x)=-inftytext{ and }lim_{xsearrow k}f(x)=+inftytag{3}$$ or $$lim_{xnearrow k}f(x)=+inftytext{ and }lim_{xsearrow k}f(x)=-infty.tag{4}$$ If $(1)$ holds, then since $f$ is continuous on $(-infty,j)$ and since $lim_{xto-infty}f(x)=1$ and $lim_{xnearrow j}f(x)=-infty,$ then $(-infty,1)$ is a subset of the range of $f,$ and so $left[-1,-frac13right]$ is, too. If $(4)$ holds, then by continuity of $f$ on $(k,infty),$ we similarly have that $left[-1,-frac13right]$ is a subset of the range of $f.$ However, if $(2)$ and $(3)$ both hold, it isn't clear from what we've discussed so far whether $left[-1,-frac13right]$ is a subset of the range of $f$ or not, so let's consider further when these different possibilities may occur.
The only way for $(1)$ to hold is if the numerator is negative at $x=j,$ meaning that $j$ must lie between the two zeroes of $x^2+x+c.$ In other words, by the quadratic formula, we have the following (equivalent) inequalities: $$frac{-1-sqrt{1^2-4c}}2<frac{-2-sqrt{2^2-4c}}2<frac{-1+sqrt{1^2-4c}}2\-1-sqrt{1-4c}<-2-sqrt{4-4c}<-1+sqrt{1-4c}\-sqrt{1-4c}<-1-sqrt{4-4c}<sqrt{1-4c}\left(-1-sqrt{4-4c}right)^2<1-4c\1+2sqrt{4-4c}+4-4c<1-4c\2sqrt{4-4c}+4<0.$$ Since this is impossible, then $(1)$ cannot hold, and so $(2)$ must hold.
Similarly, $(4)$ only holds if the numerator is negative at $x=k,$ which happens if and only if $$frac{-1-sqrt{1^2-4c}}2<frac{-2+sqrt{2^2-4c}}2<frac{-1+sqrt{1^2-4c}}2\-1-sqrt{1-4c}<-2+sqrt{4-4c}<-1+sqrt{1-4c}\-sqrt{1-4c}<-1+sqrt{4-4c}<sqrt{1-4c}\left(-1+sqrt{4-4c}right)^2<1-4c\1-2sqrt{4-4c}+4-4c<1-4c\-2sqrt{4-4c}+4<0\2<sqrt{4-4c}\4<4-4c\c<0.$$ Thus, if $c<0,$ then $left[-1,-frac13right]$ is a subset of the range of $f.$
It remains only to consider $0<c<1,$ in which case $(2)$ and $(3)$ will hold, so that $f$ achieves a local maximum in the interval $(j,k).$ Observe that $$1-c<1\sqrt{1-c}<1\2sqrt{1-c}<2\sqrt{4-4c}<2\-2+sqrt{4-4c}<0\frac{-2+sqrt{4-4c}}2<0\k<0,$$ meaning that $f$ achieves its local maximum at $x=-sqrt{c},$ and a local minimum at $x=sqrt{c}$ by the first derivative test. Since $f$ is increasing and continuous on $(-infty,j),$ and since $lim_{xto-infty}f(x)=1,$ then $f$ is positive on $(-infty,j).$ Further, since $fbigl(sqrt cbigr)$ is positive by prior work, then since $f$'s minimum value in $(k,infty)$ occurs at $x=sqrt c,$ then we have that $f$ is positive on $(k,infty).$ Thus, we need only ensure that $fleft(-sqrt cright)<-frac13.$ Since $0<c<1,$ then $c<sqrt c,$ and so the following are equivalent: $$fleft(-sqrt cright)<-frac13\-3fleft(-sqrt cright)>1\-3cdotfrac{c-sqrt c+c}{c-2sqrt c+c}>1\-3cdotfrac{2c-sqrt c}{2c-2sqrt c}>1\-3left(2c-sqrt cright)<2c-2sqrt c\-6c+3sqrt c<2c-2sqrt c\5sqrt c<8c\frac58<sqrt c\frac{25}{64}<c.$$
Putting it all together, we find that the range of $f$ does not contain the interval in question if and only if $c>frac{25}{64}.$
answered Jan 5 at 19:09
Cameron BuieCameron Buie
85.1k771155
answered Jan 5 at 19:09
Cameron BuieCameron Buie
85.1k771155
answered Jan 5 at 19:09
Cameron BuieCameron Buie
85.1k771155
answered Jan 5 at 19:09
Cameron BuieCameron Buie
85.1k771155
85.1k771155
$begingroup$
but when $c=frac12>frac{25}{64}$, it does contain the given interval, see the graph.
$endgroup$
– farruhota
Jan 5 at 19:56
1
$begingroup$
@farruhota: Incorrect. It contains points of the interval (as elements), but does not contain the interval, itself (as a subset). See the graph, yourself, and you'll find that the range is a subset of $left(-infty,-sqrt2right]cup[0,+infty),$ which does not (for example) contain $-frac13.$
$endgroup$
– Cameron Buie
Jan 5 at 20:01
$begingroup$
Agreed +1, there is a subtle difference between "completely" and "partially". In that case I must set my condition as yours $f(-sqrt{c})<-frac13$ to get the same interval for $c$ as yours. Thank you.
$endgroup$
– farruhota
Jan 5 at 20:11
$begingroup$
Even I got the same answer c>25/64. But it's wrong putting c as 26/64 f(x) is lying in the interval.
$endgroup$
– Swap Nayak
Jan 8 at 5:09
$begingroup$
@Swap See the above discussion with farruhota, as well as the discussion with John Doe. The question (as posed) is ambiguously worded.
$endgroup$
– Cameron Buie
Jan 8 at 7:53
add a comment |
$begingroup$
but when $c=frac12>frac{25}{64}$, it does contain the given interval, see the graph.
$endgroup$
– farruhota
Jan 5 at 19:56
1
$begingroup$
@farruhota: Incorrect. It contains points of the interval (as elements), but does not contain the interval, itself (as a subset). See the graph, yourself, and you'll find that the range is a subset of $left(-infty,-sqrt2right]cup[0,+infty),$ which does not (for example) contain $-frac13.$
$endgroup$
– Cameron Buie
Jan 5 at 20:01
$begingroup$
Agreed +1, there is a subtle difference between "completely" and "partially". In that case I must set my condition as yours $f(-sqrt{c})<-frac13$ to get the same interval for $c$ as yours. Thank you.
$endgroup$
– farruhota
Jan 5 at 20:11
$begingroup$
Even I got the same answer c>25/64. But it's wrong putting c as 26/64 f(x) is lying in the interval.
$endgroup$
– Swap Nayak
Jan 8 at 5:09
$begingroup$
@Swap See the above discussion with farruhota, as well as the discussion with John Doe. The question (as posed) is ambiguously worded.
$endgroup$
– Cameron Buie
Jan 8 at 7:53
$begingroup$
but when $c=frac12>frac{25}{64}$, it does contain the given interval, see the graph.
$endgroup$
– farruhota
Jan 5 at 19:56
$begingroup$
but when $c=frac12>frac{25}{64}$, it does contain the given interval, see the graph.
$endgroup$
– farruhota
Jan 5 at 19:56
1
1
$begingroup$
@farruhota: Incorrect. It contains points of the interval (as elements), but does not contain the interval, itself (as a subset). See the graph, yourself, and you'll find that the range is a subset of $left(-infty,-sqrt2right]cup[0,+infty),$ which does not (for example) contain $-frac13.$
$endgroup$
– Cameron Buie
Jan 5 at 20:01
$begingroup$
@farruhota: Incorrect. It contains points of the interval (as elements), but does not contain the interval, itself (as a subset). See the graph, yourself, and you'll find that the range is a subset of $left(-infty,-sqrt2right]cup[0,+infty),$ which does not (for example) contain $-frac13.$
$endgroup$
– Cameron Buie
Jan 5 at 20:01
$begingroup$
Agreed +1, there is a subtle difference between "completely" and "partially". In that case I must set my condition as yours $f(-sqrt{c})<-frac13$ to get the same interval for $c$ as yours. Thank you.
$endgroup$
– farruhota
Jan 5 at 20:11
$begingroup$
Agreed +1, there is a subtle difference between "completely" and "partially". In that case I must set my condition as yours $f(-sqrt{c})<-frac13$ to get the same interval for $c$ as yours. Thank you.
$endgroup$
– farruhota
Jan 5 at 20:11
$begingroup$
Even I got the same answer c>25/64. But it's wrong putting c as 26/64 f(x) is lying in the interval.
$endgroup$
– Swap Nayak
Jan 8 at 5:09
$begingroup$
Even I got the same answer c>25/64. But it's wrong putting c as 26/64 f(x) is lying in the interval.
$endgroup$
– Swap Nayak
Jan 8 at 5:09
$begingroup$
@Swap See the above discussion with farruhota, as well as the discussion with John Doe. The question (as posed) is ambiguously worded.
$endgroup$
– Cameron Buie
Jan 8 at 7:53
$begingroup$
@Swap See the above discussion with farruhota, as well as the discussion with John Doe. The question (as posed) is ambiguously worded.
$endgroup$
– Cameron Buie
Jan 8 at 7:53
add a comment |
$begingroup$
Refer to the Desmos graph.
Find the local maximum:
$$begin{align}f'(x)&= left(frac{x^2 + x + c}{x^2 + 2x + c}right)'=\
&=frac{(2x+1)(x^2+2x+c)-(x^2+x+c)(2x+2)}{(x^2 + 2x + c)^2}=0 Rightarrow \
x^2&=c Rightarrow x=pm sqrt{c}end{align}$$
Note the local maximum occurs for $x=-sqrt{c}$, for which:
$$f(-sqrt{c})=frac{2sqrt{c}-1}{2sqrt{c}-2}<-1 Rightarrow cin left(frac9{16},1right).$$
For $cin [1,+infty)$:
$$f(x)=frac{x^2+x+c}{x^2+2x+c}=1-frac{x}{x^2+2x+c}>0.$$
Hence, when $c>frac9{16}$, the function's range will not contain $[-1,-frac13]$.
answered Jan 5 at 19:51
farruhotafarruhota
19.8k2738
$endgroup$
add a comment |
$begingroup$
Refer to the Desmos graph.
Find the local maximum:
$$begin{align}f'(x)&= left(frac{x^2 + x + c}{x^2 + 2x + c}right)'=\
&=frac{(2x+1)(x^2+2x+c)-(x^2+x+c)(2x+2)}{(x^2 + 2x + c)^2}=0 Rightarrow \
x^2&=c Rightarrow x=pm sqrt{c}end{align}$$
Note the local maximum occurs for $x=-sqrt{c}$, for which:
$$f(-sqrt{c})=frac{2sqrt{c}-1}{2sqrt{c}-2}<-1 Rightarrow cin left(frac9{16},1right).$$
For $cin [1,+infty)$:
$$f(x)=frac{x^2+x+c}{x^2+2x+c}=1-frac{x}{x^2+2x+c}>0.$$
Hence, when $c>frac9{16}$, the function's range will not contain $[-1,-frac13]$.
answered Jan 5 at 19:51
farruhotafarruhota
19.8k2738
$endgroup$
add a comment |
$begingroup$
Refer to the Desmos graph.
Find the local maximum:
$$begin{align}f'(x)&= left(frac{x^2 + x + c}{x^2 + 2x + c}right)'=\
&=frac{(2x+1)(x^2+2x+c)-(x^2+x+c)(2x+2)}{(x^2 + 2x + c)^2}=0 Rightarrow \
x^2&=c Rightarrow x=pm sqrt{c}end{align}$$
Note the local maximum occurs for $x=-sqrt{c}$, for which:
$$f(-sqrt{c})=frac{2sqrt{c}-1}{2sqrt{c}-2}<-1 Rightarrow cin left(frac9{16},1right).$$
For $cin [1,+infty)$:
$$f(x)=frac{x^2+x+c}{x^2+2x+c}=1-frac{x}{x^2+2x+c}>0.$$
Hence, when $c>frac9{16}$, the function's range will not contain $[-1,-frac13]$.
answered Jan 5 at 19:51
farruhotafarruhota
19.8k2738
$endgroup$
Refer to the Desmos graph.
Find the local maximum:
$$begin{align}f'(x)&= left(frac{x^2 + x + c}{x^2 + 2x + c}right)'=\
&=frac{(2x+1)(x^2+2x+c)-(x^2+x+c)(2x+2)}{(x^2 + 2x + c)^2}=0 Rightarrow \
x^2&=c Rightarrow x=pm sqrt{c}end{align}$$
Note the local maximum occurs for $x=-sqrt{c}$, for which:
$$f(-sqrt{c})=frac{2sqrt{c}-1}{2sqrt{c}-2}<-1 Rightarrow cin left(frac9{16},1right).$$
For $cin [1,+infty)$:
$$f(x)=frac{x^2+x+c}{x^2+2x+c}=1-frac{x}{x^2+2x+c}>0.$$
Hence, when $c>frac9{16}$, the function's range will not contain $[-1,-frac13]$.
answered Jan 5 at 19:51
farruhotafarruhota
19.8k2738
answered Jan 5 at 19:51
farruhotafarruhota
19.8k2738
answered Jan 5 at 19:51
farruhotafarruhota
19.8k2738
answered Jan 5 at 19:51
farruhotafarruhota
19.8k2738
19.8k2738
add a comment |
add a comment |
$begingroup$
The inequality that you got should have been $$g(y,c)=4(1-c)y^2+2(4c-2)y+(1-4c)ge0$$Well, we want this to fail when $y$ is in this range. i.e. for $yin[-1,-1/3]$, $$4(1-c)y^2+2(4c-2)y+(1-4c)<0$$ This means if we plot this with $y$ on the $x$-axis and the value of $g(y)$ on the $y$-axis, then it will be a positive quadratic which dips below the $x$-axis for the range of $x$ values $[-1,-1/3]$. As we vary $c$, the places where this graph cuts the $x$ axis move (i.e. the roots of $g$). Note, when $c=0$, the graph is $$g(y,c=0)=4y^2-4y+1=(2y-1)^2$$which has one repeated root at $y=1/2$. Then as $c$ is increased, the minimum point of this quadratic moves leftward. The point at which $-1/3$ joins the range of values for which $g(-1/3,c)le0$ is when it first becomes a root. For what value of $c$ does this happen?
$$g(-1/3,c)=0impliescdotsimplies c=frac{25}{64}$$As we increase $c$ further, eventually $-1$ also joins this range. $$g(-1,c)=0impliescdotsimplies c=frac9{16}$$
For $c>frac9{16}$, $[-1,-1/3]$ is still not in the range of $f(x)$ - but $c=frac9{16}$ is the first value of $c$ for which this happens.
So the range of $c$ for which no points from the interval $[-1,-1/3]$ are included in the range of $f$ is given by $$bbox[5px,border:2px solid red]{cgefrac9{16}}$$
Meanwhile, the range of $c$ for which at least one point from the interval $[-1,-1/3]$ is missing from the range of $f$ is given by $$bbox[5px,border:2px solid red]{cgefrac{25}{64}}$$
answered Jan 5 at 17:35
John DoeJohn Doe
11.1k11238
$endgroup$
$begingroup$
This seems to be trying to find the values of $c$ for which the range of $f$ is disjoint from the interval in question, rather than simply not containing the interval
$endgroup$
– Cameron Buie
Jan 5 at 19:04
$begingroup$
I assumed the question was asking for the values of $c$ for which the entire interval is not in the range, but I see what you mean - it could easily also mean the range of $c$ for which any point in this interval is not in the range. In this case, this cutoff would be at $c=frac{25}{64}$, which is the point at which the point $y=-1/3$ joins the range of values for which $g(y,c)le0$.
$endgroup$
– John Doe
Jan 7 at 14:29
add a comment |
$begingroup$
The inequality that you got should have been $$g(y,c)=4(1-c)y^2+2(4c-2)y+(1-4c)ge0$$Well, we want this to fail when $y$ is in this range. i.e. for $yin[-1,-1/3]$, $$4(1-c)y^2+2(4c-2)y+(1-4c)<0$$ This means if we plot this with $y$ on the $x$-axis and the value of $g(y)$ on the $y$-axis, then it will be a positive quadratic which dips below the $x$-axis for the range of $x$ values $[-1,-1/3]$. As we vary $c$, the places where this graph cuts the $x$ axis move (i.e. the roots of $g$). Note, when $c=0$, the graph is $$g(y,c=0)=4y^2-4y+1=(2y-1)^2$$which has one repeated root at $y=1/2$. Then as $c$ is increased, the minimum point of this quadratic moves leftward. The point at which $-1/3$ joins the range of values for which $g(-1/3,c)le0$ is when it first becomes a root. For what value of $c$ does this happen?
$$g(-1/3,c)=0impliescdotsimplies c=frac{25}{64}$$As we increase $c$ further, eventually $-1$ also joins this range. $$g(-1,c)=0impliescdotsimplies c=frac9{16}$$
For $c>frac9{16}$, $[-1,-1/3]$ is still not in the range of $f(x)$ - but $c=frac9{16}$ is the first value of $c$ for which this happens.
So the range of $c$ for which no points from the interval $[-1,-1/3]$ are included in the range of $f$ is given by $$bbox[5px,border:2px solid red]{cgefrac9{16}}$$
Meanwhile, the range of $c$ for which at least one point from the interval $[-1,-1/3]$ is missing from the range of $f$ is given by $$bbox[5px,border:2px solid red]{cgefrac{25}{64}}$$
answered Jan 5 at 17:35
John DoeJohn Doe
11.1k11238
$endgroup$
$begingroup$
This seems to be trying to find the values of $c$ for which the range of $f$ is disjoint from the interval in question, rather than simply not containing the interval
$endgroup$
– Cameron Buie
Jan 5 at 19:04
$begingroup$
I assumed the question was asking for the values of $c$ for which the entire interval is not in the range, but I see what you mean - it could easily also mean the range of $c$ for which any point in this interval is not in the range. In this case, this cutoff would be at $c=frac{25}{64}$, which is the point at which the point $y=-1/3$ joins the range of values for which $g(y,c)le0$.
$endgroup$
– John Doe
Jan 7 at 14:29
add a comment |
$begingroup$
The inequality that you got should have been $$g(y,c)=4(1-c)y^2+2(4c-2)y+(1-4c)ge0$$Well, we want this to fail when $y$ is in this range. i.e. for $yin[-1,-1/3]$, $$4(1-c)y^2+2(4c-2)y+(1-4c)<0$$ This means if we plot this with $y$ on the $x$-axis and the value of $g(y)$ on the $y$-axis, then it will be a positive quadratic which dips below the $x$-axis for the range of $x$ values $[-1,-1/3]$. As we vary $c$, the places where this graph cuts the $x$ axis move (i.e. the roots of $g$). Note, when $c=0$, the graph is $$g(y,c=0)=4y^2-4y+1=(2y-1)^2$$which has one repeated root at $y=1/2$. Then as $c$ is increased, the minimum point of this quadratic moves leftward. The point at which $-1/3$ joins the range of values for which $g(-1/3,c)le0$ is when it first becomes a root. For what value of $c$ does this happen?
$$g(-1/3,c)=0impliescdotsimplies c=frac{25}{64}$$As we increase $c$ further, eventually $-1$ also joins this range. $$g(-1,c)=0impliescdotsimplies c=frac9{16}$$
For $c>frac9{16}$, $[-1,-1/3]$ is still not in the range of $f(x)$ - but $c=frac9{16}$ is the first value of $c$ for which this happens.
So the range of $c$ for which no points from the interval $[-1,-1/3]$ are included in the range of $f$ is given by $$bbox[5px,border:2px solid red]{cgefrac9{16}}$$
Meanwhile, the range of $c$ for which at least one point from the interval $[-1,-1/3]$ is missing from the range of $f$ is given by $$bbox[5px,border:2px solid red]{cgefrac{25}{64}}$$
answered Jan 5 at 17:35
John DoeJohn Doe
11.1k11238
$endgroup$
The inequality that you got should have been $$g(y,c)=4(1-c)y^2+2(4c-2)y+(1-4c)ge0$$Well, we want this to fail when $y$ is in this range. i.e. for $yin[-1,-1/3]$, $$4(1-c)y^2+2(4c-2)y+(1-4c)<0$$ This means if we plot this with $y$ on the $x$-axis and the value of $g(y)$ on the $y$-axis, then it will be a positive quadratic which dips below the $x$-axis for the range of $x$ values $[-1,-1/3]$. As we vary $c$, the places where this graph cuts the $x$ axis move (i.e. the roots of $g$). Note, when $c=0$, the graph is $$g(y,c=0)=4y^2-4y+1=(2y-1)^2$$which has one repeated root at $y=1/2$. Then as $c$ is increased, the minimum point of this quadratic moves leftward. The point at which $-1/3$ joins the range of values for which $g(-1/3,c)le0$ is when it first becomes a root. For what value of $c$ does this happen?
$$g(-1/3,c)=0impliescdotsimplies c=frac{25}{64}$$As we increase $c$ further, eventually $-1$ also joins this range. $$g(-1,c)=0impliescdotsimplies c=frac9{16}$$
For $c>frac9{16}$, $[-1,-1/3]$ is still not in the range of $f(x)$ - but $c=frac9{16}$ is the first value of $c$ for which this happens.
So the range of $c$ for which no points from the interval $[-1,-1/3]$ are included in the range of $f$ is given by $$bbox[5px,border:2px solid red]{cgefrac9{16}}$$
Meanwhile, the range of $c$ for which at least one point from the interval $[-1,-1/3]$ is missing from the range of $f$ is given by $$bbox[5px,border:2px solid red]{cgefrac{25}{64}}$$
answered Jan 5 at 17:35
John DoeJohn Doe
11.1k11238
edited Jan 7 at 14:32
answered Jan 5 at 17:35
John DoeJohn Doe
11.1k11238
answered Jan 5 at 17:35
John DoeJohn Doe
11.1k11238
answered Jan 5 at 17:35
John DoeJohn Doe
11.1k11238
11.1k11238
$begingroup$
This seems to be trying to find the values of $c$ for which the range of $f$ is disjoint from the interval in question, rather than simply not containing the interval
$endgroup$
– Cameron Buie
Jan 5 at 19:04
$begingroup$
I assumed the question was asking for the values of $c$ for which the entire interval is not in the range, but I see what you mean - it could easily also mean the range of $c$ for which any point in this interval is not in the range. In this case, this cutoff would be at $c=frac{25}{64}$, which is the point at which the point $y=-1/3$ joins the range of values for which $g(y,c)le0$.
$endgroup$
– John Doe
Jan 7 at 14:29
add a comment |
$begingroup$
This seems to be trying to find the values of $c$ for which the range of $f$ is disjoint from the interval in question, rather than simply not containing the interval
$endgroup$
– Cameron Buie
Jan 5 at 19:04
$begingroup$
I assumed the question was asking for the values of $c$ for which the entire interval is not in the range, but I see what you mean - it could easily also mean the range of $c$ for which any point in this interval is not in the range. In this case, this cutoff would be at $c=frac{25}{64}$, which is the point at which the point $y=-1/3$ joins the range of values for which $g(y,c)le0$.
$endgroup$
– John Doe
Jan 7 at 14:29
$begingroup$
This seems to be trying to find the values of $c$ for which the range of $f$ is disjoint from the interval in question, rather than simply not containing the interval
$endgroup$
– Cameron Buie
Jan 5 at 19:04
$begingroup$
This seems to be trying to find the values of $c$ for which the range of $f$ is disjoint from the interval in question, rather than simply not containing the interval
$endgroup$
– Cameron Buie
Jan 5 at 19:04
$begingroup$
I assumed the question was asking for the values of $c$ for which the entire interval is not in the range, but I see what you mean - it could easily also mean the range of $c$ for which any point in this interval is not in the range. In this case, this cutoff would be at $c=frac{25}{64}$, which is the point at which the point $y=-1/3$ joins the range of values for which $g(y,c)le0$.
$endgroup$
– John Doe
Jan 7 at 14:29
$begingroup$
I assumed the question was asking for the values of $c$ for which the entire interval is not in the range, but I see what you mean - it could easily also mean the range of $c$ for which any point in this interval is not in the range. In this case, this cutoff would be at $c=frac{25}{64}$, which is the point at which the point $y=-1/3$ joins the range of values for which $g(y,c)le0$.
$endgroup$
– John Doe
Jan 7 at 14:29
add a comment |
$begingroup$
Find the range of this function as you usually would. You should get an answer in terms of $c$. Then match this to the requirement to fix $c$.
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– John Doe
Jan 5 at 16:08
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See math.stackexchange.com/questions/2759764/finding-the-value-of-c
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– lab bhattacharjee
Jan 5 at 16:14
$begingroup$
and math.stackexchange.com/questions/1414298/…
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– lab bhattacharjee
Jan 5 at 16:15
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I let the function equal to the and got a quadratic in x. I used the fact that x is real and thus set the discriminant greater than or equal to zero. I then got a quadratic inequality in y and c. This inequality should be broken when y is between -1 and -1/3. So I graphed the new function of y and checked where the inequality was broken and calculated the corresponding value of c.
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– Swap Nayak
Jan 5 at 16:32
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But it gives the wrong answer. I guess it is because there can be multiple graphs where the inequality is broken and so multiple values of c.
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– Swap Nayak
Jan 5 at 16:35