Mixing
Range of $ y = frac{x^3+3x^2+7x-11}{x^2+5x-6} $?
$begingroup$
How do I find the range of $ y = frac{x^3+3x^2+7x-11}{x^2+5x-6} $? Would someone please help me understand how to do this by the method of forming a quadratic equation in x?
A video lesson I was watching first factorises the numerator and denominator and finds that (x-1) is a common factor. It then forms a quadratic equation in x and finds values of y for discriminant greater than or equal to zero, getting a range R. It removes the (x-1) factor from numerator and denominator. then substitutes x=1 to get a y value 16/7. I don't get why this step works. It then equates 16/7 to the original expression to find x values,one of which is one and one of which is -19/7. It concludes that range is finally R. There is also a verification step which doesn't apply to this question but to other similar questions in which we try to get coefficient of $x^2 = 0$ and then check that y value. I don't get the purpose of any of this.
I tried going through this question but it seems to deal with complex numbers and doesn't answer my doubts. What is the range of $f(x) = frac{x^2-5x+6}{x-3}$?
algebra-precalculus functions polynomials
asked Jan 13 at 2:01
HemaHema
5791113
$endgroup$
add a comment |
$begingroup$
How do I find the range of $ y = frac{x^3+3x^2+7x-11}{x^2+5x-6} $? Would someone please help me understand how to do this by the method of forming a quadratic equation in x?
A video lesson I was watching first factorises the numerator and denominator and finds that (x-1) is a common factor. It then forms a quadratic equation in x and finds values of y for discriminant greater than or equal to zero, getting a range R. It removes the (x-1) factor from numerator and denominator. then substitutes x=1 to get a y value 16/7. I don't get why this step works. It then equates 16/7 to the original expression to find x values,one of which is one and one of which is -19/7. It concludes that range is finally R. There is also a verification step which doesn't apply to this question but to other similar questions in which we try to get coefficient of $x^2 = 0$ and then check that y value. I don't get the purpose of any of this.
I tried going through this question but it seems to deal with complex numbers and doesn't answer my doubts. What is the range of $f(x) = frac{x^2-5x+6}{x-3}$?
algebra-precalculus functions polynomials
asked Jan 13 at 2:01
HemaHema
5791113
$endgroup$
2
$begingroup$
One thing to consider, since the numerator is a higher degree polynomial than the denominator, is to divide the denominator into the numerator. This is discussed, with a few good answers with detailed explanations, in Polynomial Long Division Confusion (simplifying $frac{x^{5}}{x^{2}+1}$). However, keep in mind that any values of $x$ which cause the denominator to be $0$ need to be handled appropriately.
$endgroup$
– John Omielan
Jan 13 at 2:07
1
$begingroup$
What is the result after dividing the numerator and demoninator by x - 1? Clearly 1 is not in the domain of y so calcuatling the nonexistent y(1) is TVidiotcy. At best it could be a value of a continuous extension of y.
$endgroup$
– William Elliot
Jan 13 at 3:33
add a comment |
$begingroup$
How do I find the range of $ y = frac{x^3+3x^2+7x-11}{x^2+5x-6} $? Would someone please help me understand how to do this by the method of forming a quadratic equation in x?
A video lesson I was watching first factorises the numerator and denominator and finds that (x-1) is a common factor. It then forms a quadratic equation in x and finds values of y for discriminant greater than or equal to zero, getting a range R. It removes the (x-1) factor from numerator and denominator. then substitutes x=1 to get a y value 16/7. I don't get why this step works. It then equates 16/7 to the original expression to find x values,one of which is one and one of which is -19/7. It concludes that range is finally R. There is also a verification step which doesn't apply to this question but to other similar questions in which we try to get coefficient of $x^2 = 0$ and then check that y value. I don't get the purpose of any of this.
I tried going through this question but it seems to deal with complex numbers and doesn't answer my doubts. What is the range of $f(x) = frac{x^2-5x+6}{x-3}$?
algebra-precalculus functions polynomials
asked Jan 13 at 2:01
HemaHema
5791113
$endgroup$
How do I find the range of $ y = frac{x^3+3x^2+7x-11}{x^2+5x-6} $? Would someone please help me understand how to do this by the method of forming a quadratic equation in x?
A video lesson I was watching first factorises the numerator and denominator and finds that (x-1) is a common factor. It then forms a quadratic equation in x and finds values of y for discriminant greater than or equal to zero, getting a range R. It removes the (x-1) factor from numerator and denominator. then substitutes x=1 to get a y value 16/7. I don't get why this step works. It then equates 16/7 to the original expression to find x values,one of which is one and one of which is -19/7. It concludes that range is finally R. There is also a verification step which doesn't apply to this question but to other similar questions in which we try to get coefficient of $x^2 = 0$ and then check that y value. I don't get the purpose of any of this.
I tried going through this question but it seems to deal with complex numbers and doesn't answer my doubts. What is the range of $f(x) = frac{x^2-5x+6}{x-3}$?
algebra-precalculus functions polynomials
algebra-precalculus functions polynomials
asked Jan 13 at 2:01
HemaHema
5791113
asked Jan 13 at 2:01
HemaHema
5791113
asked Jan 13 at 2:01
HemaHema
5791113
asked Jan 13 at 2:01
HemaHema
5791113
asked Jan 13 at 2:01
HemaHema
5791113
5791113
2
$begingroup$
One thing to consider, since the numerator is a higher degree polynomial than the denominator, is to divide the denominator into the numerator. This is discussed, with a few good answers with detailed explanations, in Polynomial Long Division Confusion (simplifying $frac{x^{5}}{x^{2}+1}$). However, keep in mind that any values of $x$ which cause the denominator to be $0$ need to be handled appropriately.
$endgroup$
– John Omielan
Jan 13 at 2:07
1
$begingroup$
What is the result after dividing the numerator and demoninator by x - 1? Clearly 1 is not in the domain of y so calcuatling the nonexistent y(1) is TVidiotcy. At best it could be a value of a continuous extension of y.
$endgroup$
– William Elliot
Jan 13 at 3:33
add a comment |
2
$begingroup$
One thing to consider, since the numerator is a higher degree polynomial than the denominator, is to divide the denominator into the numerator. This is discussed, with a few good answers with detailed explanations, in Polynomial Long Division Confusion (simplifying $frac{x^{5}}{x^{2}+1}$). However, keep in mind that any values of $x$ which cause the denominator to be $0$ need to be handled appropriately.
$endgroup$
– John Omielan
Jan 13 at 2:07
1
$begingroup$
What is the result after dividing the numerator and demoninator by x - 1? Clearly 1 is not in the domain of y so calcuatling the nonexistent y(1) is TVidiotcy. At best it could be a value of a continuous extension of y.
$endgroup$
– William Elliot
Jan 13 at 3:33
2
2
$begingroup$
One thing to consider, since the numerator is a higher degree polynomial than the denominator, is to divide the denominator into the numerator. This is discussed, with a few good answers with detailed explanations, in Polynomial Long Division Confusion (simplifying $frac{x^{5}}{x^{2}+1}$). However, keep in mind that any values of $x$ which cause the denominator to be $0$ need to be handled appropriately.
$endgroup$
– John Omielan
Jan 13 at 2:07
$begingroup$
One thing to consider, since the numerator is a higher degree polynomial than the denominator, is to divide the denominator into the numerator. This is discussed, with a few good answers with detailed explanations, in Polynomial Long Division Confusion (simplifying $frac{x^{5}}{x^{2}+1}$). However, keep in mind that any values of $x$ which cause the denominator to be $0$ need to be handled appropriately.
$endgroup$
– John Omielan
Jan 13 at 2:07
1
1
$begingroup$
What is the result after dividing the numerator and demoninator by x - 1? Clearly 1 is not in the domain of y so calcuatling the nonexistent y(1) is TVidiotcy. At best it could be a value of a continuous extension of y.
$endgroup$
– William Elliot
Jan 13 at 3:33
$begingroup$
What is the result after dividing the numerator and demoninator by x - 1? Clearly 1 is not in the domain of y so calcuatling the nonexistent y(1) is TVidiotcy. At best it could be a value of a continuous extension of y.
$endgroup$
– William Elliot
Jan 13 at 3:33
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$$ frac{x^2 + 4x+11}{x+6} = k ; ; , ; ; x neq 1 $$
$$ x^2 + 4x+11 = kx+6k ; ; , ; ; x neq 1 $$
$$ x^2 + (4-k)x+(11-6k) = 0 ; ; , ; ; x neq 1 $$
This has (real) solutions $x$ when the discriminant is at least zero,
$$ (4-k)^2 - 4(11-6k) geq 0, $$
$$ k^2 - 8k+ 16 +24k - 44 geq 0, $$
$$ k^2 + 16k -28 geq 0, $$
$$ k^2 + 16k +64 geq 92, $$
$$ (k+8)^2 geq 92. $$
Either
$$ k+8 geq sqrt{92} $$
or
$$ k+8 leq -sqrt{92} $$
This last pair is the same information you would get from the quadratic formula applied to $k^2 + 16 k - 28.$
I don't recall what happened with the $x neq -1,$ you should draw some graphs
Note that
$$ y = frac{x^2 + 4x+11}{x+6} $$
describes a hyperbola, vertical asymptote at $x=-6,$ the other asymptote slanted (line $y=x-2$).
answered Jan 13 at 3:25
Will JagyWill Jagy
103k5101200
$endgroup$
add a comment |
$begingroup$
We can write $$frac{x^3+3x^2+7x-11}{x^2+5x-6}=frac{x^2+4x+11}{x+6}=x-2+frac{23}{x+6},quad xne 1.
$$ If $x+6>0$, then by AM-GM inequality, we have
$$
x-2+frac{23}{x+6}=(x+6)+frac{23}{x+6}-8ge -8+2sqrt{23}.
$$ (Equality holds for $x+6=sqrt{23}$.) We can observe that the value of the function taken by $x+6=7$ is also attained by $x+6=frac{23}{7}$. Since it holds that $$lim_{xtoinfty}left(x-2+frac{23}{x+6}right)=lim_{xto -6^+}left(x-2+frac{23}{x+6}right)=infty,$$ the intermediate value theorem implies that the range of the given function on ${x+6>0, xne 1}$ is $$[-8+2sqrt{23},infty).$$
On the other hand, if $x+6<0$, then
$$
(-x-6)+left(-frac{23}{x+6}right)ge 2sqrt{23}
$$ and
$$
(x+6)+frac{23}{x+6}-8le -8-2sqrt{23}.
$$ Since it holds $lim_{xto-infty}left(x-2+frac{23}{x+6}right)=-infty$, the range on ${x+6<0}$ is given by $$(-infty,-8-2sqrt{23}]$$ also by the intermediate value theorem.
Gathering them, the range of the given function is
$$
(-infty,-8-2sqrt{23}]cup[-8+2sqrt{23},infty).
$$
answered Jan 13 at 6:17
SongSong
12.6k631
$endgroup$
add a comment |
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2 Answers
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$begingroup$
$$ frac{x^2 + 4x+11}{x+6} = k ; ; , ; ; x neq 1 $$
$$ x^2 + 4x+11 = kx+6k ; ; , ; ; x neq 1 $$
$$ x^2 + (4-k)x+(11-6k) = 0 ; ; , ; ; x neq 1 $$
This has (real) solutions $x$ when the discriminant is at least zero,
$$ (4-k)^2 - 4(11-6k) geq 0, $$
$$ k^2 - 8k+ 16 +24k - 44 geq 0, $$
$$ k^2 + 16k -28 geq 0, $$
$$ k^2 + 16k +64 geq 92, $$
$$ (k+8)^2 geq 92. $$
Either
$$ k+8 geq sqrt{92} $$
or
$$ k+8 leq -sqrt{92} $$
This last pair is the same information you would get from the quadratic formula applied to $k^2 + 16 k - 28.$
I don't recall what happened with the $x neq -1,$ you should draw some graphs
Note that
$$ y = frac{x^2 + 4x+11}{x+6} $$
describes a hyperbola, vertical asymptote at $x=-6,$ the other asymptote slanted (line $y=x-2$).
answered Jan 13 at 3:25
Will JagyWill Jagy
103k5101200
$endgroup$
add a comment |
$begingroup$
$$ frac{x^2 + 4x+11}{x+6} = k ; ; , ; ; x neq 1 $$
$$ x^2 + 4x+11 = kx+6k ; ; , ; ; x neq 1 $$
$$ x^2 + (4-k)x+(11-6k) = 0 ; ; , ; ; x neq 1 $$
This has (real) solutions $x$ when the discriminant is at least zero,
$$ (4-k)^2 - 4(11-6k) geq 0, $$
$$ k^2 - 8k+ 16 +24k - 44 geq 0, $$
$$ k^2 + 16k -28 geq 0, $$
$$ k^2 + 16k +64 geq 92, $$
$$ (k+8)^2 geq 92. $$
Either
$$ k+8 geq sqrt{92} $$
or
$$ k+8 leq -sqrt{92} $$
This last pair is the same information you would get from the quadratic formula applied to $k^2 + 16 k - 28.$
I don't recall what happened with the $x neq -1,$ you should draw some graphs
Note that
$$ y = frac{x^2 + 4x+11}{x+6} $$
describes a hyperbola, vertical asymptote at $x=-6,$ the other asymptote slanted (line $y=x-2$).
answered Jan 13 at 3:25
Will JagyWill Jagy
103k5101200
$endgroup$
add a comment |
$begingroup$
$$ frac{x^2 + 4x+11}{x+6} = k ; ; , ; ; x neq 1 $$
$$ x^2 + 4x+11 = kx+6k ; ; , ; ; x neq 1 $$
$$ x^2 + (4-k)x+(11-6k) = 0 ; ; , ; ; x neq 1 $$
This has (real) solutions $x$ when the discriminant is at least zero,
$$ (4-k)^2 - 4(11-6k) geq 0, $$
$$ k^2 - 8k+ 16 +24k - 44 geq 0, $$
$$ k^2 + 16k -28 geq 0, $$
$$ k^2 + 16k +64 geq 92, $$
$$ (k+8)^2 geq 92. $$
Either
$$ k+8 geq sqrt{92} $$
or
$$ k+8 leq -sqrt{92} $$
This last pair is the same information you would get from the quadratic formula applied to $k^2 + 16 k - 28.$
I don't recall what happened with the $x neq -1,$ you should draw some graphs
Note that
$$ y = frac{x^2 + 4x+11}{x+6} $$
describes a hyperbola, vertical asymptote at $x=-6,$ the other asymptote slanted (line $y=x-2$).
answered Jan 13 at 3:25
Will JagyWill Jagy
103k5101200
$endgroup$
$$ frac{x^2 + 4x+11}{x+6} = k ; ; , ; ; x neq 1 $$
$$ x^2 + 4x+11 = kx+6k ; ; , ; ; x neq 1 $$
$$ x^2 + (4-k)x+(11-6k) = 0 ; ; , ; ; x neq 1 $$
This has (real) solutions $x$ when the discriminant is at least zero,
$$ (4-k)^2 - 4(11-6k) geq 0, $$
$$ k^2 - 8k+ 16 +24k - 44 geq 0, $$
$$ k^2 + 16k -28 geq 0, $$
$$ k^2 + 16k +64 geq 92, $$
$$ (k+8)^2 geq 92. $$
Either
$$ k+8 geq sqrt{92} $$
or
$$ k+8 leq -sqrt{92} $$
This last pair is the same information you would get from the quadratic formula applied to $k^2 + 16 k - 28.$
I don't recall what happened with the $x neq -1,$ you should draw some graphs
Note that
$$ y = frac{x^2 + 4x+11}{x+6} $$
describes a hyperbola, vertical asymptote at $x=-6,$ the other asymptote slanted (line $y=x-2$).
answered Jan 13 at 3:25
Will JagyWill Jagy
103k5101200
edited Jan 13 at 3:35
answered Jan 13 at 3:25
Will JagyWill Jagy
103k5101200
answered Jan 13 at 3:25
Will JagyWill Jagy
103k5101200
answered Jan 13 at 3:25
Will JagyWill Jagy
103k5101200
103k5101200
add a comment |
add a comment |
$begingroup$
We can write $$frac{x^3+3x^2+7x-11}{x^2+5x-6}=frac{x^2+4x+11}{x+6}=x-2+frac{23}{x+6},quad xne 1.
$$ If $x+6>0$, then by AM-GM inequality, we have
$$
x-2+frac{23}{x+6}=(x+6)+frac{23}{x+6}-8ge -8+2sqrt{23}.
$$ (Equality holds for $x+6=sqrt{23}$.) We can observe that the value of the function taken by $x+6=7$ is also attained by $x+6=frac{23}{7}$. Since it holds that $$lim_{xtoinfty}left(x-2+frac{23}{x+6}right)=lim_{xto -6^+}left(x-2+frac{23}{x+6}right)=infty,$$ the intermediate value theorem implies that the range of the given function on ${x+6>0, xne 1}$ is $$[-8+2sqrt{23},infty).$$
On the other hand, if $x+6<0$, then
$$
(-x-6)+left(-frac{23}{x+6}right)ge 2sqrt{23}
$$ and
$$
(x+6)+frac{23}{x+6}-8le -8-2sqrt{23}.
$$ Since it holds $lim_{xto-infty}left(x-2+frac{23}{x+6}right)=-infty$, the range on ${x+6<0}$ is given by $$(-infty,-8-2sqrt{23}]$$ also by the intermediate value theorem.
Gathering them, the range of the given function is
$$
(-infty,-8-2sqrt{23}]cup[-8+2sqrt{23},infty).
$$
answered Jan 13 at 6:17
SongSong
12.6k631
$endgroup$
add a comment |
$begingroup$
We can write $$frac{x^3+3x^2+7x-11}{x^2+5x-6}=frac{x^2+4x+11}{x+6}=x-2+frac{23}{x+6},quad xne 1.
$$ If $x+6>0$, then by AM-GM inequality, we have
$$
x-2+frac{23}{x+6}=(x+6)+frac{23}{x+6}-8ge -8+2sqrt{23}.
$$ (Equality holds for $x+6=sqrt{23}$.) We can observe that the value of the function taken by $x+6=7$ is also attained by $x+6=frac{23}{7}$. Since it holds that $$lim_{xtoinfty}left(x-2+frac{23}{x+6}right)=lim_{xto -6^+}left(x-2+frac{23}{x+6}right)=infty,$$ the intermediate value theorem implies that the range of the given function on ${x+6>0, xne 1}$ is $$[-8+2sqrt{23},infty).$$
On the other hand, if $x+6<0$, then
$$
(-x-6)+left(-frac{23}{x+6}right)ge 2sqrt{23}
$$ and
$$
(x+6)+frac{23}{x+6}-8le -8-2sqrt{23}.
$$ Since it holds $lim_{xto-infty}left(x-2+frac{23}{x+6}right)=-infty$, the range on ${x+6<0}$ is given by $$(-infty,-8-2sqrt{23}]$$ also by the intermediate value theorem.
Gathering them, the range of the given function is
$$
(-infty,-8-2sqrt{23}]cup[-8+2sqrt{23},infty).
$$
answered Jan 13 at 6:17
SongSong
12.6k631
$endgroup$
add a comment |
$begingroup$
We can write $$frac{x^3+3x^2+7x-11}{x^2+5x-6}=frac{x^2+4x+11}{x+6}=x-2+frac{23}{x+6},quad xne 1.
$$ If $x+6>0$, then by AM-GM inequality, we have
$$
x-2+frac{23}{x+6}=(x+6)+frac{23}{x+6}-8ge -8+2sqrt{23}.
$$ (Equality holds for $x+6=sqrt{23}$.) We can observe that the value of the function taken by $x+6=7$ is also attained by $x+6=frac{23}{7}$. Since it holds that $$lim_{xtoinfty}left(x-2+frac{23}{x+6}right)=lim_{xto -6^+}left(x-2+frac{23}{x+6}right)=infty,$$ the intermediate value theorem implies that the range of the given function on ${x+6>0, xne 1}$ is $$[-8+2sqrt{23},infty).$$
On the other hand, if $x+6<0$, then
$$
(-x-6)+left(-frac{23}{x+6}right)ge 2sqrt{23}
$$ and
$$
(x+6)+frac{23}{x+6}-8le -8-2sqrt{23}.
$$ Since it holds $lim_{xto-infty}left(x-2+frac{23}{x+6}right)=-infty$, the range on ${x+6<0}$ is given by $$(-infty,-8-2sqrt{23}]$$ also by the intermediate value theorem.
Gathering them, the range of the given function is
$$
(-infty,-8-2sqrt{23}]cup[-8+2sqrt{23},infty).
$$
answered Jan 13 at 6:17
SongSong
12.6k631
$endgroup$
We can write $$frac{x^3+3x^2+7x-11}{x^2+5x-6}=frac{x^2+4x+11}{x+6}=x-2+frac{23}{x+6},quad xne 1.
$$ If $x+6>0$, then by AM-GM inequality, we have
$$
x-2+frac{23}{x+6}=(x+6)+frac{23}{x+6}-8ge -8+2sqrt{23}.
$$ (Equality holds for $x+6=sqrt{23}$.) We can observe that the value of the function taken by $x+6=7$ is also attained by $x+6=frac{23}{7}$. Since it holds that $$lim_{xtoinfty}left(x-2+frac{23}{x+6}right)=lim_{xto -6^+}left(x-2+frac{23}{x+6}right)=infty,$$ the intermediate value theorem implies that the range of the given function on ${x+6>0, xne 1}$ is $$[-8+2sqrt{23},infty).$$
On the other hand, if $x+6<0$, then
$$
(-x-6)+left(-frac{23}{x+6}right)ge 2sqrt{23}
$$ and
$$
(x+6)+frac{23}{x+6}-8le -8-2sqrt{23}.
$$ Since it holds $lim_{xto-infty}left(x-2+frac{23}{x+6}right)=-infty$, the range on ${x+6<0}$ is given by $$(-infty,-8-2sqrt{23}]$$ also by the intermediate value theorem.
Gathering them, the range of the given function is
$$
(-infty,-8-2sqrt{23}]cup[-8+2sqrt{23},infty).
$$
answered Jan 13 at 6:17
SongSong
12.6k631
edited Jan 13 at 6:46
answered Jan 13 at 6:17
SongSong
12.6k631
answered Jan 13 at 6:17
SongSong
12.6k631
answered Jan 13 at 6:17
SongSong
12.6k631
12.6k631
add a comment |
add a comment |
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2
$begingroup$
One thing to consider, since the numerator is a higher degree polynomial than the denominator, is to divide the denominator into the numerator. This is discussed, with a few good answers with detailed explanations, in Polynomial Long Division Confusion (simplifying $frac{x^{5}}{x^{2}+1}$). However, keep in mind that any values of $x$ which cause the denominator to be $0$ need to be handled appropriately.
$endgroup$
– John Omielan
Jan 13 at 2:07
1
$begingroup$
What is the result after dividing the numerator and demoninator by x - 1? Clearly 1 is not in the domain of y so calcuatling the nonexistent y(1) is TVidiotcy. At best it could be a value of a continuous extension of y.
$endgroup$
– William Elliot
Jan 13 at 3:33