Mixing
$f(x) = (x-2)(x-4)(x-6) +2$ then $f$ has all real roots between $0$ and $6$. True or false?
$begingroup$
$f(x) = (x-2)(x-4)(x-6) +2$ then $f$ has all real roots between $0$ and $6$
$($ true or false$)?$
Here
$f(0) = -46$ and $f(6) = 2$ since function is continuous so it must have at least one root between $0$ and $6$, but how to check if it has all its roots between $0$ and $6$, without really finding out the roots?
calculus
asked Jan 22 at 3:37
MathsaddictMathsaddict
3669
$endgroup$
|
show 1 more comment
$begingroup$
$f(x) = (x-2)(x-4)(x-6) +2$ then $f$ has all real roots between $0$ and $6$
$($ true or false$)?$
Here
$f(0) = -46$ and $f(6) = 2$ since function is continuous so it must have at least one root between $0$ and $6$, but how to check if it has all its roots between $0$ and $6$, without really finding out the roots?
calculus
asked Jan 22 at 3:37
MathsaddictMathsaddict
3669
$endgroup$
$begingroup$
As the question asks whether or not all real roots are between $6$ and $7$, finding even one real root outside of that range, as you have done, shows whether or not the requested statement is true. You don't need to show all roots are between $0$ and $6$ or any other such range. Or is the issue possibly due to a typo and it's to check for all real roots are between $0$ and $6$ instead?
$endgroup$
– John Omielan
Jan 22 at 3:43
$begingroup$
We can check that the $f'(x)>0$ for all $x<0$ as well as $x>6$. Since $f(0)<0$, $f(x)<0$ for $x<0$. Similarly, $f(x)>0$ for $x>6$.
$endgroup$
– Kelvin Soh
Jan 22 at 3:43
$begingroup$
Hint: what is $f(2)$?
$endgroup$
– J. W. Tanner
Jan 22 at 3:45
$begingroup$
$f(2)=2$ and also $f(5)=-1$ then we are done.
$endgroup$
– Offlaw
Jan 22 at 3:46
$begingroup$
I'm really sorry . I edited the question, it is to find if f has all its roots between 0 and 6 or not.
$endgroup$
– Mathsaddict
Jan 22 at 3:51
|
show 1 more comment
$begingroup$
$f(x) = (x-2)(x-4)(x-6) +2$ then $f$ has all real roots between $0$ and $6$
$($ true or false$)?$
Here
$f(0) = -46$ and $f(6) = 2$ since function is continuous so it must have at least one root between $0$ and $6$, but how to check if it has all its roots between $0$ and $6$, without really finding out the roots?
calculus
asked Jan 22 at 3:37
MathsaddictMathsaddict
3669
$endgroup$
$f(x) = (x-2)(x-4)(x-6) +2$ then $f$ has all real roots between $0$ and $6$
$($ true or false$)?$
Here
$f(0) = -46$ and $f(6) = 2$ since function is continuous so it must have at least one root between $0$ and $6$, but how to check if it has all its roots between $0$ and $6$, without really finding out the roots?
calculus
calculus
asked Jan 22 at 3:37
MathsaddictMathsaddict
3669
asked Jan 22 at 3:37
MathsaddictMathsaddict
3669
edited Jan 22 at 3:50
Mathsaddict
asked Jan 22 at 3:37
MathsaddictMathsaddict
3669
asked Jan 22 at 3:37
MathsaddictMathsaddict
3669
asked Jan 22 at 3:37
MathsaddictMathsaddict
3669
3669
$begingroup$
As the question asks whether or not all real roots are between $6$ and $7$, finding even one real root outside of that range, as you have done, shows whether or not the requested statement is true. You don't need to show all roots are between $0$ and $6$ or any other such range. Or is the issue possibly due to a typo and it's to check for all real roots are between $0$ and $6$ instead?
$endgroup$
– John Omielan
Jan 22 at 3:43
$begingroup$
We can check that the $f'(x)>0$ for all $x<0$ as well as $x>6$. Since $f(0)<0$, $f(x)<0$ for $x<0$. Similarly, $f(x)>0$ for $x>6$.
$endgroup$
– Kelvin Soh
Jan 22 at 3:43
$begingroup$
Hint: what is $f(2)$?
$endgroup$
– J. W. Tanner
Jan 22 at 3:45
$begingroup$
$f(2)=2$ and also $f(5)=-1$ then we are done.
$endgroup$
– Offlaw
Jan 22 at 3:46
$begingroup$
I'm really sorry . I edited the question, it is to find if f has all its roots between 0 and 6 or not.
$endgroup$
– Mathsaddict
Jan 22 at 3:51
|
show 1 more comment
$begingroup$
As the question asks whether or not all real roots are between $6$ and $7$, finding even one real root outside of that range, as you have done, shows whether or not the requested statement is true. You don't need to show all roots are between $0$ and $6$ or any other such range. Or is the issue possibly due to a typo and it's to check for all real roots are between $0$ and $6$ instead?
$endgroup$
– John Omielan
Jan 22 at 3:43
$begingroup$
We can check that the $f'(x)>0$ for all $x<0$ as well as $x>6$. Since $f(0)<0$, $f(x)<0$ for $x<0$. Similarly, $f(x)>0$ for $x>6$.
$endgroup$
– Kelvin Soh
Jan 22 at 3:43
$begingroup$
Hint: what is $f(2)$?
$endgroup$
– J. W. Tanner
Jan 22 at 3:45
$begingroup$
$f(2)=2$ and also $f(5)=-1$ then we are done.
$endgroup$
– Offlaw
Jan 22 at 3:46
$begingroup$
I'm really sorry . I edited the question, it is to find if f has all its roots between 0 and 6 or not.
$endgroup$
– Mathsaddict
Jan 22 at 3:51
$begingroup$
As the question asks whether or not all real roots are between $6$ and $7$, finding even one real root outside of that range, as you have done, shows whether or not the requested statement is true. You don't need to show all roots are between $0$ and $6$ or any other such range. Or is the issue possibly due to a typo and it's to check for all real roots are between $0$ and $6$ instead?
$endgroup$
– John Omielan
Jan 22 at 3:43
$begingroup$
As the question asks whether or not all real roots are between $6$ and $7$, finding even one real root outside of that range, as you have done, shows whether or not the requested statement is true. You don't need to show all roots are between $0$ and $6$ or any other such range. Or is the issue possibly due to a typo and it's to check for all real roots are between $0$ and $6$ instead?
$endgroup$
– John Omielan
Jan 22 at 3:43
$begingroup$
We can check that the $f'(x)>0$ for all $x<0$ as well as $x>6$. Since $f(0)<0$, $f(x)<0$ for $x<0$. Similarly, $f(x)>0$ for $x>6$.
$endgroup$
– Kelvin Soh
Jan 22 at 3:43
$begingroup$
We can check that the $f'(x)>0$ for all $x<0$ as well as $x>6$. Since $f(0)<0$, $f(x)<0$ for $x<0$. Similarly, $f(x)>0$ for $x>6$.
$endgroup$
– Kelvin Soh
Jan 22 at 3:43
$begingroup$
Hint: what is $f(2)$?
$endgroup$
– J. W. Tanner
Jan 22 at 3:45
$begingroup$
Hint: what is $f(2)$?
$endgroup$
– J. W. Tanner
Jan 22 at 3:45
$begingroup$
$f(2)=2$ and also $f(5)=-1$ then we are done.
$endgroup$
– Offlaw
Jan 22 at 3:46
$begingroup$
$f(2)=2$ and also $f(5)=-1$ then we are done.
$endgroup$
– Offlaw
Jan 22 at 3:46
$begingroup$
I'm really sorry . I edited the question, it is to find if f has all its roots between 0 and 6 or not.
$endgroup$
– Mathsaddict
Jan 22 at 3:51
$begingroup$
I'm really sorry . I edited the question, it is to find if f has all its roots between 0 and 6 or not.
$endgroup$
– Mathsaddict
Jan 22 at 3:51
|
show 1 more comment
4 Answers
4
active
oldest
votes
$begingroup$
Check $$f(1),f(2)$$ and $$f(4),f(5)$$ and $$f(5),f(6)$$
answered Jan 22 at 4:11
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.2k42866
$endgroup$
$begingroup$
$f(1)=-13, f(2) = 2, f(4) = 2, f(5) = -1, f(6)=2$ so, $f$ has three real roots between 0 and 6. And f has almost 3 real roots, so all roots are in between 0 and 6. How would I estimate these points, where signs of function are alternatively positive and negative to solve such problems?
$endgroup$
– Mathsaddict
Jan 22 at 4:19
1
$begingroup$
@Mathsaddict: without the $+2$ the roots are at $2,4,6$. The big worry is that the root at $6$ will shift outside the interval. As the graph is rising through $6$ on its way to $+infty$ lifting it up by $2$ will move the root to the left. Points halfway between the old roots are good candidates to not switch signs, so evaluating at $1,2,3,4,5,6$ is a good way to go.
$endgroup$
– Ross Millikan
Jan 22 at 4:27
add a comment |
$begingroup$
Well,
$$alpha >0 to f(-alpha)=2-(alpha+2)(alpha+4)(alpha+6) < -46$$
and
$$f(6+alpha)=2+(4+alpha)(2+alpha)(alpha)> 2$$
So at the very least all its real roots are $in (0,6)$
You just need to show all of its roots are real.
answered Jan 22 at 4:17
Rhys HughesRhys Hughes
6,9741530
$endgroup$
1
$begingroup$
The statement "has all real roots between $0$ and $6$" can be interpreted in $2$ ways. One is that all of the roots are real and in that range, and the other is that any roots which are real (e.g., perhaps there is only $1$ real root) are in that range. When I first read that statement, I assumed the latter case. I believe the question could be worded better to avoid this potential ambiguity.
$endgroup$
– John Omielan
Jan 22 at 4:23
$begingroup$
@JohnOmielan you misquoted me, and thusly caused your own confusion. I said all of the function's real roots were in $(0,6)$, not that the function has all real roots in $(0,6)$
$endgroup$
– Rhys Hughes
Jan 22 at 4:42
$begingroup$
Actually, I was quoting from the question, not what you wrote. My interpretation was that you just had to show that any roots which are real are in $left(0,6right)$, which you showed in the first part of your answer. However, if you interpret the question as stating that all roots are real and are in the $left(0,6right)$ range, then your last sentence makes sense. It's a relatively minor point which I perhaps shouldn't have even mentioned so, if you wish, please ignore this and I can delete my comments.
$endgroup$
– John Omielan
Jan 22 at 4:48
add a comment |
$begingroup$
In this problem, you don't have to actually calculate the roots (as you pointed out). One way you could solve this is to use the intermediate value theorem (going off of Dr. Sonnhard Graubner's comment):
f(1) = -13
f(2) = 2
By the intermediate value theorem, you know that f has to pass through 0 at some point between f(1) and f(2), meaning that one of its roots is between x=1 and x=2, fulfilling the requirement.
f(4) = 2
f(5) = -1
Same process as above - by the intermediate value theorem, you know there must be another root between x=4 and x=5.
f(6) = 2
You also know there must be a root between x=5 and x=6.
answered Jan 22 at 4:30
elbeckerelbecker
33
$endgroup$
add a comment |
$begingroup$
Consider $F(x):=f(x)-2= (x-2)(x-4)(x-6).$
1)$F(x), f(x)$ polynomials of degree $3.$
2) $x rightarrow infty$ $F(x), f(x) rightarrow infty.$
3) $x rightarrow -infty$ $F(x), f(x) rightarrow -infty.$
4) Roots of $F(x)$ at $x=2,4,6$. That's it.
5) $x> 6:$ $F(x) > 0,$ $f(x)= F(x)+2$.
No roots of $f$ for $x>6$.
6) For $x <2$, $F(x) <0$, and
$F(x)= -(2-x)(4-x)(6-x)$ is strictly decreasing.
7)Likewise for $x <2$ $f(x)= F(x)+2$ is strictly decreasing
$F(1)=-(1)(3)(5)=-15$, and $f(1)= -15+2=-13$.
No root of $f$ for $x <1$.
Hence?
answered Jan 22 at 10:21
Peter SzilasPeter Szilas
11.5k2822
$endgroup$
add a comment |
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4 Answers
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active
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4 Answers
4
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$begingroup$
Check $$f(1),f(2)$$ and $$f(4),f(5)$$ and $$f(5),f(6)$$
answered Jan 22 at 4:11
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.2k42866
$endgroup$
$begingroup$
$f(1)=-13, f(2) = 2, f(4) = 2, f(5) = -1, f(6)=2$ so, $f$ has three real roots between 0 and 6. And f has almost 3 real roots, so all roots are in between 0 and 6. How would I estimate these points, where signs of function are alternatively positive and negative to solve such problems?
$endgroup$
– Mathsaddict
Jan 22 at 4:19
1
$begingroup$
@Mathsaddict: without the $+2$ the roots are at $2,4,6$. The big worry is that the root at $6$ will shift outside the interval. As the graph is rising through $6$ on its way to $+infty$ lifting it up by $2$ will move the root to the left. Points halfway between the old roots are good candidates to not switch signs, so evaluating at $1,2,3,4,5,6$ is a good way to go.
$endgroup$
– Ross Millikan
Jan 22 at 4:27
add a comment |
$begingroup$
Check $$f(1),f(2)$$ and $$f(4),f(5)$$ and $$f(5),f(6)$$
answered Jan 22 at 4:11
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.2k42866
$endgroup$
$begingroup$
$f(1)=-13, f(2) = 2, f(4) = 2, f(5) = -1, f(6)=2$ so, $f$ has three real roots between 0 and 6. And f has almost 3 real roots, so all roots are in between 0 and 6. How would I estimate these points, where signs of function are alternatively positive and negative to solve such problems?
$endgroup$
– Mathsaddict
Jan 22 at 4:19
1
$begingroup$
@Mathsaddict: without the $+2$ the roots are at $2,4,6$. The big worry is that the root at $6$ will shift outside the interval. As the graph is rising through $6$ on its way to $+infty$ lifting it up by $2$ will move the root to the left. Points halfway between the old roots are good candidates to not switch signs, so evaluating at $1,2,3,4,5,6$ is a good way to go.
$endgroup$
– Ross Millikan
Jan 22 at 4:27
add a comment |
$begingroup$
Check $$f(1),f(2)$$ and $$f(4),f(5)$$ and $$f(5),f(6)$$
answered Jan 22 at 4:11
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.2k42866
$endgroup$
Check $$f(1),f(2)$$ and $$f(4),f(5)$$ and $$f(5),f(6)$$
answered Jan 22 at 4:11
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.2k42866
answered Jan 22 at 4:11
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.2k42866
answered Jan 22 at 4:11
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.2k42866
answered Jan 22 at 4:11
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
77.2k42866
77.2k42866
$begingroup$
$f(1)=-13, f(2) = 2, f(4) = 2, f(5) = -1, f(6)=2$ so, $f$ has three real roots between 0 and 6. And f has almost 3 real roots, so all roots are in between 0 and 6. How would I estimate these points, where signs of function are alternatively positive and negative to solve such problems?
$endgroup$
– Mathsaddict
Jan 22 at 4:19
1
$begingroup$
@Mathsaddict: without the $+2$ the roots are at $2,4,6$. The big worry is that the root at $6$ will shift outside the interval. As the graph is rising through $6$ on its way to $+infty$ lifting it up by $2$ will move the root to the left. Points halfway between the old roots are good candidates to not switch signs, so evaluating at $1,2,3,4,5,6$ is a good way to go.
$endgroup$
– Ross Millikan
Jan 22 at 4:27
add a comment |
$begingroup$
$f(1)=-13, f(2) = 2, f(4) = 2, f(5) = -1, f(6)=2$ so, $f$ has three real roots between 0 and 6. And f has almost 3 real roots, so all roots are in between 0 and 6. How would I estimate these points, where signs of function are alternatively positive and negative to solve such problems?
$endgroup$
– Mathsaddict
Jan 22 at 4:19
1
$begingroup$
@Mathsaddict: without the $+2$ the roots are at $2,4,6$. The big worry is that the root at $6$ will shift outside the interval. As the graph is rising through $6$ on its way to $+infty$ lifting it up by $2$ will move the root to the left. Points halfway between the old roots are good candidates to not switch signs, so evaluating at $1,2,3,4,5,6$ is a good way to go.
$endgroup$
– Ross Millikan
Jan 22 at 4:27
$begingroup$
$f(1)=-13, f(2) = 2, f(4) = 2, f(5) = -1, f(6)=2$ so, $f$ has three real roots between 0 and 6. And f has almost 3 real roots, so all roots are in between 0 and 6. How would I estimate these points, where signs of function are alternatively positive and negative to solve such problems?
$endgroup$
– Mathsaddict
Jan 22 at 4:19
$begingroup$
$f(1)=-13, f(2) = 2, f(4) = 2, f(5) = -1, f(6)=2$ so, $f$ has three real roots between 0 and 6. And f has almost 3 real roots, so all roots are in between 0 and 6. How would I estimate these points, where signs of function are alternatively positive and negative to solve such problems?
$endgroup$
– Mathsaddict
Jan 22 at 4:19
1
1
$begingroup$
@Mathsaddict: without the $+2$ the roots are at $2,4,6$. The big worry is that the root at $6$ will shift outside the interval. As the graph is rising through $6$ on its way to $+infty$ lifting it up by $2$ will move the root to the left. Points halfway between the old roots are good candidates to not switch signs, so evaluating at $1,2,3,4,5,6$ is a good way to go.
$endgroup$
– Ross Millikan
Jan 22 at 4:27
$begingroup$
@Mathsaddict: without the $+2$ the roots are at $2,4,6$. The big worry is that the root at $6$ will shift outside the interval. As the graph is rising through $6$ on its way to $+infty$ lifting it up by $2$ will move the root to the left. Points halfway between the old roots are good candidates to not switch signs, so evaluating at $1,2,3,4,5,6$ is a good way to go.
$endgroup$
– Ross Millikan
Jan 22 at 4:27
add a comment |
$begingroup$
Well,
$$alpha >0 to f(-alpha)=2-(alpha+2)(alpha+4)(alpha+6) < -46$$
and
$$f(6+alpha)=2+(4+alpha)(2+alpha)(alpha)> 2$$
So at the very least all its real roots are $in (0,6)$
You just need to show all of its roots are real.
answered Jan 22 at 4:17
Rhys HughesRhys Hughes
6,9741530
$endgroup$
1
$begingroup$
The statement "has all real roots between $0$ and $6$" can be interpreted in $2$ ways. One is that all of the roots are real and in that range, and the other is that any roots which are real (e.g., perhaps there is only $1$ real root) are in that range. When I first read that statement, I assumed the latter case. I believe the question could be worded better to avoid this potential ambiguity.
$endgroup$
– John Omielan
Jan 22 at 4:23
$begingroup$
@JohnOmielan you misquoted me, and thusly caused your own confusion. I said all of the function's real roots were in $(0,6)$, not that the function has all real roots in $(0,6)$
$endgroup$
– Rhys Hughes
Jan 22 at 4:42
$begingroup$
Actually, I was quoting from the question, not what you wrote. My interpretation was that you just had to show that any roots which are real are in $left(0,6right)$, which you showed in the first part of your answer. However, if you interpret the question as stating that all roots are real and are in the $left(0,6right)$ range, then your last sentence makes sense. It's a relatively minor point which I perhaps shouldn't have even mentioned so, if you wish, please ignore this and I can delete my comments.
$endgroup$
– John Omielan
Jan 22 at 4:48
add a comment |
$begingroup$
Well,
$$alpha >0 to f(-alpha)=2-(alpha+2)(alpha+4)(alpha+6) < -46$$
and
$$f(6+alpha)=2+(4+alpha)(2+alpha)(alpha)> 2$$
So at the very least all its real roots are $in (0,6)$
You just need to show all of its roots are real.
answered Jan 22 at 4:17
Rhys HughesRhys Hughes
6,9741530
$endgroup$
1
$begingroup$
The statement "has all real roots between $0$ and $6$" can be interpreted in $2$ ways. One is that all of the roots are real and in that range, and the other is that any roots which are real (e.g., perhaps there is only $1$ real root) are in that range. When I first read that statement, I assumed the latter case. I believe the question could be worded better to avoid this potential ambiguity.
$endgroup$
– John Omielan
Jan 22 at 4:23
$begingroup$
@JohnOmielan you misquoted me, and thusly caused your own confusion. I said all of the function's real roots were in $(0,6)$, not that the function has all real roots in $(0,6)$
$endgroup$
– Rhys Hughes
Jan 22 at 4:42
$begingroup$
Actually, I was quoting from the question, not what you wrote. My interpretation was that you just had to show that any roots which are real are in $left(0,6right)$, which you showed in the first part of your answer. However, if you interpret the question as stating that all roots are real and are in the $left(0,6right)$ range, then your last sentence makes sense. It's a relatively minor point which I perhaps shouldn't have even mentioned so, if you wish, please ignore this and I can delete my comments.
$endgroup$
– John Omielan
Jan 22 at 4:48
add a comment |
$begingroup$
Well,
$$alpha >0 to f(-alpha)=2-(alpha+2)(alpha+4)(alpha+6) < -46$$
and
$$f(6+alpha)=2+(4+alpha)(2+alpha)(alpha)> 2$$
So at the very least all its real roots are $in (0,6)$
You just need to show all of its roots are real.
answered Jan 22 at 4:17
Rhys HughesRhys Hughes
6,9741530
$endgroup$
Well,
$$alpha >0 to f(-alpha)=2-(alpha+2)(alpha+4)(alpha+6) < -46$$
and
$$f(6+alpha)=2+(4+alpha)(2+alpha)(alpha)> 2$$
So at the very least all its real roots are $in (0,6)$
You just need to show all of its roots are real.
answered Jan 22 at 4:17
Rhys HughesRhys Hughes
6,9741530
answered Jan 22 at 4:17
Rhys HughesRhys Hughes
6,9741530
answered Jan 22 at 4:17
Rhys HughesRhys Hughes
6,9741530
answered Jan 22 at 4:17
Rhys HughesRhys Hughes
6,9741530
6,9741530
1
$begingroup$
The statement "has all real roots between $0$ and $6$" can be interpreted in $2$ ways. One is that all of the roots are real and in that range, and the other is that any roots which are real (e.g., perhaps there is only $1$ real root) are in that range. When I first read that statement, I assumed the latter case. I believe the question could be worded better to avoid this potential ambiguity.
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– John Omielan
Jan 22 at 4:23
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@JohnOmielan you misquoted me, and thusly caused your own confusion. I said all of the function's real roots were in $(0,6)$, not that the function has all real roots in $(0,6)$
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– Rhys Hughes
Jan 22 at 4:42
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Actually, I was quoting from the question, not what you wrote. My interpretation was that you just had to show that any roots which are real are in $left(0,6right)$, which you showed in the first part of your answer. However, if you interpret the question as stating that all roots are real and are in the $left(0,6right)$ range, then your last sentence makes sense. It's a relatively minor point which I perhaps shouldn't have even mentioned so, if you wish, please ignore this and I can delete my comments.
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– John Omielan
Jan 22 at 4:48
add a comment |
1
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The statement "has all real roots between $0$ and $6$" can be interpreted in $2$ ways. One is that all of the roots are real and in that range, and the other is that any roots which are real (e.g., perhaps there is only $1$ real root) are in that range. When I first read that statement, I assumed the latter case. I believe the question could be worded better to avoid this potential ambiguity.
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– John Omielan
Jan 22 at 4:23
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@JohnOmielan you misquoted me, and thusly caused your own confusion. I said all of the function's real roots were in $(0,6)$, not that the function has all real roots in $(0,6)$
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– Rhys Hughes
Jan 22 at 4:42
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Actually, I was quoting from the question, not what you wrote. My interpretation was that you just had to show that any roots which are real are in $left(0,6right)$, which you showed in the first part of your answer. However, if you interpret the question as stating that all roots are real and are in the $left(0,6right)$ range, then your last sentence makes sense. It's a relatively minor point which I perhaps shouldn't have even mentioned so, if you wish, please ignore this and I can delete my comments.
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– John Omielan
Jan 22 at 4:48
1
1
$begingroup$
The statement "has all real roots between $0$ and $6$" can be interpreted in $2$ ways. One is that all of the roots are real and in that range, and the other is that any roots which are real (e.g., perhaps there is only $1$ real root) are in that range. When I first read that statement, I assumed the latter case. I believe the question could be worded better to avoid this potential ambiguity.
$endgroup$
– John Omielan
Jan 22 at 4:23
$begingroup$
The statement "has all real roots between $0$ and $6$" can be interpreted in $2$ ways. One is that all of the roots are real and in that range, and the other is that any roots which are real (e.g., perhaps there is only $1$ real root) are in that range. When I first read that statement, I assumed the latter case. I believe the question could be worded better to avoid this potential ambiguity.
$endgroup$
– John Omielan
Jan 22 at 4:23
$begingroup$
@JohnOmielan you misquoted me, and thusly caused your own confusion. I said all of the function's real roots were in $(0,6)$, not that the function has all real roots in $(0,6)$
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– Rhys Hughes
Jan 22 at 4:42
$begingroup$
@JohnOmielan you misquoted me, and thusly caused your own confusion. I said all of the function's real roots were in $(0,6)$, not that the function has all real roots in $(0,6)$
$endgroup$
– Rhys Hughes
Jan 22 at 4:42
$begingroup$
Actually, I was quoting from the question, not what you wrote. My interpretation was that you just had to show that any roots which are real are in $left(0,6right)$, which you showed in the first part of your answer. However, if you interpret the question as stating that all roots are real and are in the $left(0,6right)$ range, then your last sentence makes sense. It's a relatively minor point which I perhaps shouldn't have even mentioned so, if you wish, please ignore this and I can delete my comments.
$endgroup$
– John Omielan
Jan 22 at 4:48
$begingroup$
Actually, I was quoting from the question, not what you wrote. My interpretation was that you just had to show that any roots which are real are in $left(0,6right)$, which you showed in the first part of your answer. However, if you interpret the question as stating that all roots are real and are in the $left(0,6right)$ range, then your last sentence makes sense. It's a relatively minor point which I perhaps shouldn't have even mentioned so, if you wish, please ignore this and I can delete my comments.
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– John Omielan
Jan 22 at 4:48
add a comment |
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In this problem, you don't have to actually calculate the roots (as you pointed out). One way you could solve this is to use the intermediate value theorem (going off of Dr. Sonnhard Graubner's comment):
f(1) = -13
f(2) = 2
By the intermediate value theorem, you know that f has to pass through 0 at some point between f(1) and f(2), meaning that one of its roots is between x=1 and x=2, fulfilling the requirement.
f(4) = 2
f(5) = -1
Same process as above - by the intermediate value theorem, you know there must be another root between x=4 and x=5.
f(6) = 2
You also know there must be a root between x=5 and x=6.
answered Jan 22 at 4:30
elbeckerelbecker
33
$endgroup$
add a comment |
$begingroup$
In this problem, you don't have to actually calculate the roots (as you pointed out). One way you could solve this is to use the intermediate value theorem (going off of Dr. Sonnhard Graubner's comment):
f(1) = -13
f(2) = 2
By the intermediate value theorem, you know that f has to pass through 0 at some point between f(1) and f(2), meaning that one of its roots is between x=1 and x=2, fulfilling the requirement.
f(4) = 2
f(5) = -1
Same process as above - by the intermediate value theorem, you know there must be another root between x=4 and x=5.
f(6) = 2
You also know there must be a root between x=5 and x=6.
answered Jan 22 at 4:30
elbeckerelbecker
33
$endgroup$
add a comment |
$begingroup$
In this problem, you don't have to actually calculate the roots (as you pointed out). One way you could solve this is to use the intermediate value theorem (going off of Dr. Sonnhard Graubner's comment):
f(1) = -13
f(2) = 2
By the intermediate value theorem, you know that f has to pass through 0 at some point between f(1) and f(2), meaning that one of its roots is between x=1 and x=2, fulfilling the requirement.
f(4) = 2
f(5) = -1
Same process as above - by the intermediate value theorem, you know there must be another root between x=4 and x=5.
f(6) = 2
You also know there must be a root between x=5 and x=6.
answered Jan 22 at 4:30
elbeckerelbecker
33
$endgroup$
In this problem, you don't have to actually calculate the roots (as you pointed out). One way you could solve this is to use the intermediate value theorem (going off of Dr. Sonnhard Graubner's comment):
f(1) = -13
f(2) = 2
By the intermediate value theorem, you know that f has to pass through 0 at some point between f(1) and f(2), meaning that one of its roots is between x=1 and x=2, fulfilling the requirement.
f(4) = 2
f(5) = -1
Same process as above - by the intermediate value theorem, you know there must be another root between x=4 and x=5.
f(6) = 2
You also know there must be a root between x=5 and x=6.
answered Jan 22 at 4:30
elbeckerelbecker
33
edited Jan 22 at 4:35
answered Jan 22 at 4:30
elbeckerelbecker
33
answered Jan 22 at 4:30
elbeckerelbecker
33
answered Jan 22 at 4:30
elbeckerelbecker
33
33
add a comment |
add a comment |
$begingroup$
Consider $F(x):=f(x)-2= (x-2)(x-4)(x-6).$
1)$F(x), f(x)$ polynomials of degree $3.$
2) $x rightarrow infty$ $F(x), f(x) rightarrow infty.$
3) $x rightarrow -infty$ $F(x), f(x) rightarrow -infty.$
4) Roots of $F(x)$ at $x=2,4,6$. That's it.
5) $x> 6:$ $F(x) > 0,$ $f(x)= F(x)+2$.
No roots of $f$ for $x>6$.
6) For $x <2$, $F(x) <0$, and
$F(x)= -(2-x)(4-x)(6-x)$ is strictly decreasing.
7)Likewise for $x <2$ $f(x)= F(x)+2$ is strictly decreasing
$F(1)=-(1)(3)(5)=-15$, and $f(1)= -15+2=-13$.
No root of $f$ for $x <1$.
Hence?
answered Jan 22 at 10:21
Peter SzilasPeter Szilas
11.5k2822
$endgroup$
add a comment |
$begingroup$
Consider $F(x):=f(x)-2= (x-2)(x-4)(x-6).$
1)$F(x), f(x)$ polynomials of degree $3.$
2) $x rightarrow infty$ $F(x), f(x) rightarrow infty.$
3) $x rightarrow -infty$ $F(x), f(x) rightarrow -infty.$
4) Roots of $F(x)$ at $x=2,4,6$. That's it.
5) $x> 6:$ $F(x) > 0,$ $f(x)= F(x)+2$.
No roots of $f$ for $x>6$.
6) For $x <2$, $F(x) <0$, and
$F(x)= -(2-x)(4-x)(6-x)$ is strictly decreasing.
7)Likewise for $x <2$ $f(x)= F(x)+2$ is strictly decreasing
$F(1)=-(1)(3)(5)=-15$, and $f(1)= -15+2=-13$.
No root of $f$ for $x <1$.
Hence?
answered Jan 22 at 10:21
Peter SzilasPeter Szilas
11.5k2822
$endgroup$
add a comment |
$begingroup$
Consider $F(x):=f(x)-2= (x-2)(x-4)(x-6).$
1)$F(x), f(x)$ polynomials of degree $3.$
2) $x rightarrow infty$ $F(x), f(x) rightarrow infty.$
3) $x rightarrow -infty$ $F(x), f(x) rightarrow -infty.$
4) Roots of $F(x)$ at $x=2,4,6$. That's it.
5) $x> 6:$ $F(x) > 0,$ $f(x)= F(x)+2$.
No roots of $f$ for $x>6$.
6) For $x <2$, $F(x) <0$, and
$F(x)= -(2-x)(4-x)(6-x)$ is strictly decreasing.
7)Likewise for $x <2$ $f(x)= F(x)+2$ is strictly decreasing
$F(1)=-(1)(3)(5)=-15$, and $f(1)= -15+2=-13$.
No root of $f$ for $x <1$.
Hence?
answered Jan 22 at 10:21
Peter SzilasPeter Szilas
11.5k2822
$endgroup$
Consider $F(x):=f(x)-2= (x-2)(x-4)(x-6).$
1)$F(x), f(x)$ polynomials of degree $3.$
2) $x rightarrow infty$ $F(x), f(x) rightarrow infty.$
3) $x rightarrow -infty$ $F(x), f(x) rightarrow -infty.$
4) Roots of $F(x)$ at $x=2,4,6$. That's it.
5) $x> 6:$ $F(x) > 0,$ $f(x)= F(x)+2$.
No roots of $f$ for $x>6$.
6) For $x <2$, $F(x) <0$, and
$F(x)= -(2-x)(4-x)(6-x)$ is strictly decreasing.
7)Likewise for $x <2$ $f(x)= F(x)+2$ is strictly decreasing
$F(1)=-(1)(3)(5)=-15$, and $f(1)= -15+2=-13$.
No root of $f$ for $x <1$.
Hence?
answered Jan 22 at 10:21
Peter SzilasPeter Szilas
11.5k2822
edited Jan 22 at 10:26
answered Jan 22 at 10:21
Peter SzilasPeter Szilas
11.5k2822
answered Jan 22 at 10:21
Peter SzilasPeter Szilas
11.5k2822
answered Jan 22 at 10:21
Peter SzilasPeter Szilas
11.5k2822
11.5k2822
add a comment |
add a comment |
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$begingroup$
As the question asks whether or not all real roots are between $6$ and $7$, finding even one real root outside of that range, as you have done, shows whether or not the requested statement is true. You don't need to show all roots are between $0$ and $6$ or any other such range. Or is the issue possibly due to a typo and it's to check for all real roots are between $0$ and $6$ instead?
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– John Omielan
Jan 22 at 3:43
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We can check that the $f'(x)>0$ for all $x<0$ as well as $x>6$. Since $f(0)<0$, $f(x)<0$ for $x<0$. Similarly, $f(x)>0$ for $x>6$.
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– Kelvin Soh
Jan 22 at 3:43
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Hint: what is $f(2)$?
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– J. W. Tanner
Jan 22 at 3:45
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$f(2)=2$ and also $f(5)=-1$ then we are done.
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– Offlaw
Jan 22 at 3:46
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I'm really sorry . I edited the question, it is to find if f has all its roots between 0 and 6 or not.
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– Mathsaddict
Jan 22 at 3:51