X-coordinates of critical points are independent of m and n
$begingroup$
I was given a function $f(x)=a(-3x+2)^7-b,$ where a and b are constants that do not equal zero. I am supposed to Show that the x-coordinate(s) of the location(s) of any critical points are independent of a and b. I took the first derivative and second derivative already. I am stuck on how to show this. Thank you!
calculus derivatives
asked Jan 15 at 23:04
math lovermath lover
33
$endgroup$
|
show 1 more comment
$begingroup$
I was given a function $f(x)=a(-3x+2)^7-b,$ where a and b are constants that do not equal zero. I am supposed to Show that the x-coordinate(s) of the location(s) of any critical points are independent of a and b. I took the first derivative and second derivative already. I am stuck on how to show this. Thank you!
calculus derivatives
asked Jan 15 at 23:04
math lovermath lover
33
$endgroup$
$begingroup$
By critical point, do you mean a point at which the derivative is zero? I.e. a stationary point? Also, what are $m$ and $n$? You haven't said what they are anywhere, so surely any result you get is independent of them?
$endgroup$
– John Doe
Jan 15 at 23:14
$begingroup$
yes, a critical point is a place where the derivative is und or zero.
$endgroup$
– math lover
Jan 15 at 23:16
$begingroup$
What does "und" mean?
$endgroup$
– William Elliot
Jan 15 at 23:30
$begingroup$
it means undefined
$endgroup$
– math lover
Jan 15 at 23:37
$begingroup$
what did you get for the first derivative when you worked it out? You are supposed to set this equal to $0$, and solve for $x$. I am pretty sure that this should give you a solution for $x$ that does not depend on $a$ or $b$, which is what the question asks you to show.
$endgroup$
– John Doe
Jan 17 at 2:25
|
show 1 more comment
$begingroup$
I was given a function $f(x)=a(-3x+2)^7-b,$ where a and b are constants that do not equal zero. I am supposed to Show that the x-coordinate(s) of the location(s) of any critical points are independent of a and b. I took the first derivative and second derivative already. I am stuck on how to show this. Thank you!
calculus derivatives
asked Jan 15 at 23:04
math lovermath lover
33
$endgroup$
I was given a function $f(x)=a(-3x+2)^7-b,$ where a and b are constants that do not equal zero. I am supposed to Show that the x-coordinate(s) of the location(s) of any critical points are independent of a and b. I took the first derivative and second derivative already. I am stuck on how to show this. Thank you!
calculus derivatives
calculus derivatives
asked Jan 15 at 23:04
math lovermath lover
33
asked Jan 15 at 23:04
math lovermath lover
33
edited Jan 16 at 3:39
math lover
asked Jan 15 at 23:04
math lovermath lover
33
asked Jan 15 at 23:04
math lovermath lover
33
asked Jan 15 at 23:04
math lovermath lover
33
33
$begingroup$
By critical point, do you mean a point at which the derivative is zero? I.e. a stationary point? Also, what are $m$ and $n$? You haven't said what they are anywhere, so surely any result you get is independent of them?
$endgroup$
– John Doe
Jan 15 at 23:14
$begingroup$
yes, a critical point is a place where the derivative is und or zero.
$endgroup$
– math lover
Jan 15 at 23:16
$begingroup$
What does "und" mean?
$endgroup$
– William Elliot
Jan 15 at 23:30
$begingroup$
it means undefined
$endgroup$
– math lover
Jan 15 at 23:37
$begingroup$
what did you get for the first derivative when you worked it out? You are supposed to set this equal to $0$, and solve for $x$. I am pretty sure that this should give you a solution for $x$ that does not depend on $a$ or $b$, which is what the question asks you to show.
$endgroup$
– John Doe
Jan 17 at 2:25
|
show 1 more comment
$begingroup$
By critical point, do you mean a point at which the derivative is zero? I.e. a stationary point? Also, what are $m$ and $n$? You haven't said what they are anywhere, so surely any result you get is independent of them?
$endgroup$
– John Doe
Jan 15 at 23:14
$begingroup$
yes, a critical point is a place where the derivative is und or zero.
$endgroup$
– math lover
Jan 15 at 23:16
$begingroup$
What does "und" mean?
$endgroup$
– William Elliot
Jan 15 at 23:30
$begingroup$
it means undefined
$endgroup$
– math lover
Jan 15 at 23:37
$begingroup$
what did you get for the first derivative when you worked it out? You are supposed to set this equal to $0$, and solve for $x$. I am pretty sure that this should give you a solution for $x$ that does not depend on $a$ or $b$, which is what the question asks you to show.
$endgroup$
– John Doe
Jan 17 at 2:25
$begingroup$
By critical point, do you mean a point at which the derivative is zero? I.e. a stationary point? Also, what are $m$ and $n$? You haven't said what they are anywhere, so surely any result you get is independent of them?
$endgroup$
– John Doe
Jan 15 at 23:14
$begingroup$
By critical point, do you mean a point at which the derivative is zero? I.e. a stationary point? Also, what are $m$ and $n$? You haven't said what they are anywhere, so surely any result you get is independent of them?
$endgroup$
– John Doe
Jan 15 at 23:14
$begingroup$
yes, a critical point is a place where the derivative is und or zero.
$endgroup$
– math lover
Jan 15 at 23:16
$begingroup$
yes, a critical point is a place where the derivative is und or zero.
$endgroup$
– math lover
Jan 15 at 23:16
$begingroup$
What does "und" mean?
$endgroup$
– William Elliot
Jan 15 at 23:30
$begingroup$
What does "und" mean?
$endgroup$
– William Elliot
Jan 15 at 23:30
$begingroup$
it means undefined
$endgroup$
– math lover
Jan 15 at 23:37
$begingroup$
it means undefined
$endgroup$
– math lover
Jan 15 at 23:37
$begingroup$
what did you get for the first derivative when you worked it out? You are supposed to set this equal to $0$, and solve for $x$. I am pretty sure that this should give you a solution for $x$ that does not depend on $a$ or $b$, which is what the question asks you to show.
$endgroup$
– John Doe
Jan 17 at 2:25
$begingroup$
what did you get for the first derivative when you worked it out? You are supposed to set this equal to $0$, and solve for $x$. I am pretty sure that this should give you a solution for $x$ that does not depend on $a$ or $b$, which is what the question asks you to show.
$endgroup$
– John Doe
Jan 17 at 2:25
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
$f'(x) = -3a$ is very dependent upon $a$ and independent of $b$.
There are no critical points, points x for which $f'(x) = 0$, as $a$ is nonzero.
It makes no difference if $b$ is zero.
So as there are no critical points, their location,
namely nowhere, is independent of any nonzero $a$.
By the power of the lucky $7$,
$f'(x) = -21ax(-3x + 2)^6 = 0 $
which for nonzero $a$ is equivalent to $x(-3x + 2)^6 = 0. $
So the critical points are independent of $b$ and nonzero $a$.
Why second derivatives?
Are you looking for local minimums/maximums, inflection points?
edited Jan 17 at 13:53
John Doe
12.2k11340
answered Jan 15 at 23:41
William ElliotWilliam Elliot
9,2912820
$endgroup$
$begingroup$
Hey! The question has been edited. It was missing a power of $7$, so now it probably can have turning points.
$endgroup$
– John Doe
Jan 17 at 2:23
$begingroup$
@JohnDoe. See edit.
$endgroup$
– William Elliot
Jan 17 at 7:06
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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votes
$begingroup$
$f'(x) = -3a$ is very dependent upon $a$ and independent of $b$.
There are no critical points, points x for which $f'(x) = 0$, as $a$ is nonzero.
It makes no difference if $b$ is zero.
So as there are no critical points, their location,
namely nowhere, is independent of any nonzero $a$.
By the power of the lucky $7$,
$f'(x) = -21ax(-3x + 2)^6 = 0 $
which for nonzero $a$ is equivalent to $x(-3x + 2)^6 = 0. $
So the critical points are independent of $b$ and nonzero $a$.
Why second derivatives?
Are you looking for local minimums/maximums, inflection points?
edited Jan 17 at 13:53
John Doe
12.2k11340
answered Jan 15 at 23:41
William ElliotWilliam Elliot
9,2912820
$endgroup$
$begingroup$
Hey! The question has been edited. It was missing a power of $7$, so now it probably can have turning points.
$endgroup$
– John Doe
Jan 17 at 2:23
$begingroup$
@JohnDoe. See edit.
$endgroup$
– William Elliot
Jan 17 at 7:06
add a comment |
$begingroup$
$f'(x) = -3a$ is very dependent upon $a$ and independent of $b$.
There are no critical points, points x for which $f'(x) = 0$, as $a$ is nonzero.
It makes no difference if $b$ is zero.
So as there are no critical points, their location,
namely nowhere, is independent of any nonzero $a$.
By the power of the lucky $7$,
$f'(x) = -21ax(-3x + 2)^6 = 0 $
which for nonzero $a$ is equivalent to $x(-3x + 2)^6 = 0. $
So the critical points are independent of $b$ and nonzero $a$.
Why second derivatives?
Are you looking for local minimums/maximums, inflection points?
edited Jan 17 at 13:53
John Doe
12.2k11340
answered Jan 15 at 23:41
William ElliotWilliam Elliot
9,2912820
$endgroup$
$begingroup$
Hey! The question has been edited. It was missing a power of $7$, so now it probably can have turning points.
$endgroup$
– John Doe
Jan 17 at 2:23
$begingroup$
@JohnDoe. See edit.
$endgroup$
– William Elliot
Jan 17 at 7:06
add a comment |
$begingroup$
$f'(x) = -3a$ is very dependent upon $a$ and independent of $b$.
There are no critical points, points x for which $f'(x) = 0$, as $a$ is nonzero.
It makes no difference if $b$ is zero.
So as there are no critical points, their location,
namely nowhere, is independent of any nonzero $a$.
By the power of the lucky $7$,
$f'(x) = -21ax(-3x + 2)^6 = 0 $
which for nonzero $a$ is equivalent to $x(-3x + 2)^6 = 0. $
So the critical points are independent of $b$ and nonzero $a$.
Why second derivatives?
Are you looking for local minimums/maximums, inflection points?
edited Jan 17 at 13:53
John Doe
12.2k11340
answered Jan 15 at 23:41
William ElliotWilliam Elliot
9,2912820
$endgroup$
$f'(x) = -3a$ is very dependent upon $a$ and independent of $b$.
There are no critical points, points x for which $f'(x) = 0$, as $a$ is nonzero.
It makes no difference if $b$ is zero.
So as there are no critical points, their location,
namely nowhere, is independent of any nonzero $a$.
By the power of the lucky $7$,
$f'(x) = -21ax(-3x + 2)^6 = 0 $
which for nonzero $a$ is equivalent to $x(-3x + 2)^6 = 0. $
So the critical points are independent of $b$ and nonzero $a$.
Why second derivatives?
Are you looking for local minimums/maximums, inflection points?
edited Jan 17 at 13:53
John Doe
12.2k11340
answered Jan 15 at 23:41
William ElliotWilliam Elliot
9,2912820
edited Jan 17 at 13:53
John Doe
12.2k11340
edited Jan 17 at 13:53
John Doe
12.2k11340
edited Jan 17 at 13:53
John Doe
12.2k11340
12.2k11340
answered Jan 15 at 23:41
William ElliotWilliam Elliot
9,2912820
answered Jan 15 at 23:41
William ElliotWilliam Elliot
9,2912820
answered Jan 15 at 23:41
William ElliotWilliam Elliot
9,2912820
9,2912820
$begingroup$
Hey! The question has been edited. It was missing a power of $7$, so now it probably can have turning points.
$endgroup$
– John Doe
Jan 17 at 2:23
$begingroup$
@JohnDoe. See edit.
$endgroup$
– William Elliot
Jan 17 at 7:06
add a comment |
$begingroup$
Hey! The question has been edited. It was missing a power of $7$, so now it probably can have turning points.
$endgroup$
– John Doe
Jan 17 at 2:23
$begingroup$
@JohnDoe. See edit.
$endgroup$
– William Elliot
Jan 17 at 7:06
$begingroup$
Hey! The question has been edited. It was missing a power of $7$, so now it probably can have turning points.
$endgroup$
– John Doe
Jan 17 at 2:23
$begingroup$
Hey! The question has been edited. It was missing a power of $7$, so now it probably can have turning points.
$endgroup$
– John Doe
Jan 17 at 2:23
$begingroup$
@JohnDoe. See edit.
$endgroup$
– William Elliot
Jan 17 at 7:06
$begingroup$
@JohnDoe. See edit.
$endgroup$
– William Elliot
Jan 17 at 7:06
add a comment |
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$begingroup$
By critical point, do you mean a point at which the derivative is zero? I.e. a stationary point? Also, what are $m$ and $n$? You haven't said what they are anywhere, so surely any result you get is independent of them?
$endgroup$
– John Doe
Jan 15 at 23:14
$begingroup$
yes, a critical point is a place where the derivative is und or zero.
$endgroup$
– math lover
Jan 15 at 23:16
$begingroup$
What does "und" mean?
$endgroup$
– William Elliot
Jan 15 at 23:30
$begingroup$
it means undefined
$endgroup$
– math lover
Jan 15 at 23:37
$begingroup$
what did you get for the first derivative when you worked it out? You are supposed to set this equal to $0$, and solve for $x$. I am pretty sure that this should give you a solution for $x$ that does not depend on $a$ or $b$, which is what the question asks you to show.
$endgroup$
– John Doe
Jan 17 at 2:25