Find all values of $c$ for the Mean Value Theorem's $f'(c)$ of the function $f(x) = 2x^3-6x^2-90x+6$ of the range $[-5, 8]$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)For what values of $b$ does this function lack extrema?“Box With No Top” OptimizationA way of finding the range of an injective functionFinding the ideal angle for the maximum range of a projectile when elevated?Finding total work by integrationFind the value of constants $c_1, c_2, c_3, c_4$ for which function $f: mathbbR rightarrow mathbbR$ is differentiableFind the maximum and minimum values of the functionFind shortest collection of roads connecting 4 towns at 4 corners of rectangle with lengths 1 and 'a' for sides of rectanglefinding the range of values for a constant such that a certain function does not take the value zeroFind $y'$ for the equation: $xe^-y/2 + ye^-x/2 = sin(x^2+y^2)$
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Find all values of $c$ for the Mean Value Theorem's $f'(c)$ of the function $f(x) = 2x^3-6x^2-90x+6$ of the range $[-5, 8]$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)For what values of $b$ does this function lack extrema?“Box With No Top” OptimizationA way of finding the range of an injective functionFinding the ideal angle for the maximum range of a projectile when elevated?Finding total work by integrationFind the value of constants $c_1, c_2, c_3, c_4$ for which function $f: mathbbR rightarrow mathbbR$ is differentiableFind the maximum and minimum values of the functionFind shortest collection of roads connecting 4 towns at 4 corners of rectangle with lengths 1 and 'a' for sides of rectanglefinding the range of values for a constant such that a certain function does not take the value zeroFind $y'$ for the equation: $xe^-y/2 + ye^-x/2 = sin(x^2+y^2)$
$begingroup$
I need to find all values of $c$ for the $f'(c)$ of the function $f(x) = 2x^3-6x^2-90x+6$ of the range $[-5, 8]$ such that $f'(c) = -10$. I have already found that $f'(c) = frac-13013=-10$.
I thought I could find the values of $c$ by differentiating $f(x)$ which becomes $6x^2-12x-90$, and applying the Mean Value Theorem $f'(c) = fracf(b)-f(a)b - a$ again.
$f(b) = 6(-5)^2-12(-5)-90 = 150+60-90 = 120$
$f(a) = 6(8)^2-12(8)-90 = 384-186 = 198$
$f'(c) = frac120-198-5-8 = frac7813 = 6$
Then I took $6x^2-12x-90$ again, replaced $x$ with $c$, and set it equal to $6$ before solving for $c$.
$6c^2-12c-90 = 6 to 6(c^2-2c-16) = 0$
$frac2 pm sqrt2^2-4(1)(-17)2(1) = frac2 pm sqrt742$ or $c = frac2 - sqrt742, frac2 + sqrt742$
But this answer is wrong.
How can I find the values for $c$ here?
calculus derivatives
asked Mar 31 at 22:15
LuminousNutriaLuminousNutria
57612
$endgroup$
add a comment |
$begingroup$
I need to find all values of $c$ for the $f'(c)$ of the function $f(x) = 2x^3-6x^2-90x+6$ of the range $[-5, 8]$ such that $f'(c) = -10$. I have already found that $f'(c) = frac-13013=-10$.
I thought I could find the values of $c$ by differentiating $f(x)$ which becomes $6x^2-12x-90$, and applying the Mean Value Theorem $f'(c) = fracf(b)-f(a)b - a$ again.
$f(b) = 6(-5)^2-12(-5)-90 = 150+60-90 = 120$
$f(a) = 6(8)^2-12(8)-90 = 384-186 = 198$
$f'(c) = frac120-198-5-8 = frac7813 = 6$
Then I took $6x^2-12x-90$ again, replaced $x$ with $c$, and set it equal to $6$ before solving for $c$.
$6c^2-12c-90 = 6 to 6(c^2-2c-16) = 0$
$frac2 pm sqrt2^2-4(1)(-17)2(1) = frac2 pm sqrt742$ or $c = frac2 - sqrt742, frac2 + sqrt742$
But this answer is wrong.
How can I find the values for $c$ here?
calculus derivatives
asked Mar 31 at 22:15
LuminousNutriaLuminousNutria
57612
$endgroup$
2
$begingroup$
Are you trying to find all $cin [-5,8]$ such that $f'(c) = fracf(8)-f(-5)8-(-5)$? Your question is not clear.
$endgroup$
– rogerl
Mar 31 at 22:18
$begingroup$
You evaluated $f'(b)$, not $f(b)$ (and same for $a$)
$endgroup$
– J. W. Tanner
Mar 31 at 22:20
$begingroup$
@rogerl I don't understand your math symbol $epsilon$ (I only know it's name). I have applied the Mean Value Theorem to $f(x) = 2x^3-6x^2-90x+6$ to find out $f'(c) = frac-13013$ now I need to find all values of $c$ such that $f'(c) = frac-13013$
$endgroup$
– LuminousNutria
Mar 31 at 22:21
1
$begingroup$
$in$ means "is contained in" (the interval)
$endgroup$
– J. W. Tanner
Mar 31 at 22:22
add a comment |
$begingroup$
I need to find all values of $c$ for the $f'(c)$ of the function $f(x) = 2x^3-6x^2-90x+6$ of the range $[-5, 8]$ such that $f'(c) = -10$. I have already found that $f'(c) = frac-13013=-10$.
I thought I could find the values of $c$ by differentiating $f(x)$ which becomes $6x^2-12x-90$, and applying the Mean Value Theorem $f'(c) = fracf(b)-f(a)b - a$ again.
$f(b) = 6(-5)^2-12(-5)-90 = 150+60-90 = 120$
$f(a) = 6(8)^2-12(8)-90 = 384-186 = 198$
$f'(c) = frac120-198-5-8 = frac7813 = 6$
Then I took $6x^2-12x-90$ again, replaced $x$ with $c$, and set it equal to $6$ before solving for $c$.
$6c^2-12c-90 = 6 to 6(c^2-2c-16) = 0$
$frac2 pm sqrt2^2-4(1)(-17)2(1) = frac2 pm sqrt742$ or $c = frac2 - sqrt742, frac2 + sqrt742$
But this answer is wrong.
How can I find the values for $c$ here?
calculus derivatives
asked Mar 31 at 22:15
LuminousNutriaLuminousNutria
57612
$endgroup$
I need to find all values of $c$ for the $f'(c)$ of the function $f(x) = 2x^3-6x^2-90x+6$ of the range $[-5, 8]$ such that $f'(c) = -10$. I have already found that $f'(c) = frac-13013=-10$.
I thought I could find the values of $c$ by differentiating $f(x)$ which becomes $6x^2-12x-90$, and applying the Mean Value Theorem $f'(c) = fracf(b)-f(a)b - a$ again.
$f(b) = 6(-5)^2-12(-5)-90 = 150+60-90 = 120$
$f(a) = 6(8)^2-12(8)-90 = 384-186 = 198$
$f'(c) = frac120-198-5-8 = frac7813 = 6$
Then I took $6x^2-12x-90$ again, replaced $x$ with $c$, and set it equal to $6$ before solving for $c$.
$6c^2-12c-90 = 6 to 6(c^2-2c-16) = 0$
$frac2 pm sqrt2^2-4(1)(-17)2(1) = frac2 pm sqrt742$ or $c = frac2 - sqrt742, frac2 + sqrt742$
But this answer is wrong.
How can I find the values for $c$ here?
calculus derivatives
calculus derivatives
asked Mar 31 at 22:15
LuminousNutriaLuminousNutria
57612
asked Mar 31 at 22:15
LuminousNutriaLuminousNutria
57612
edited Mar 31 at 22:22
LuminousNutria
asked Mar 31 at 22:15
LuminousNutriaLuminousNutria
57612
asked Mar 31 at 22:15
LuminousNutriaLuminousNutria
57612
asked Mar 31 at 22:15
LuminousNutriaLuminousNutria
57612
57612
2
$begingroup$
Are you trying to find all $cin [-5,8]$ such that $f'(c) = fracf(8)-f(-5)8-(-5)$? Your question is not clear.
$endgroup$
– rogerl
Mar 31 at 22:18
$begingroup$
You evaluated $f'(b)$, not $f(b)$ (and same for $a$)
$endgroup$
– J. W. Tanner
Mar 31 at 22:20
$begingroup$
@rogerl I don't understand your math symbol $epsilon$ (I only know it's name). I have applied the Mean Value Theorem to $f(x) = 2x^3-6x^2-90x+6$ to find out $f'(c) = frac-13013$ now I need to find all values of $c$ such that $f'(c) = frac-13013$
$endgroup$
– LuminousNutria
Mar 31 at 22:21
1
$begingroup$
$in$ means "is contained in" (the interval)
$endgroup$
– J. W. Tanner
Mar 31 at 22:22
add a comment |
2
$begingroup$
Are you trying to find all $cin [-5,8]$ such that $f'(c) = fracf(8)-f(-5)8-(-5)$? Your question is not clear.
$endgroup$
– rogerl
Mar 31 at 22:18
$begingroup$
You evaluated $f'(b)$, not $f(b)$ (and same for $a$)
$endgroup$
– J. W. Tanner
Mar 31 at 22:20
$begingroup$
@rogerl I don't understand your math symbol $epsilon$ (I only know it's name). I have applied the Mean Value Theorem to $f(x) = 2x^3-6x^2-90x+6$ to find out $f'(c) = frac-13013$ now I need to find all values of $c$ such that $f'(c) = frac-13013$
$endgroup$
– LuminousNutria
Mar 31 at 22:21
1
$begingroup$
$in$ means "is contained in" (the interval)
$endgroup$
– J. W. Tanner
Mar 31 at 22:22
2
2
$begingroup$
Are you trying to find all $cin [-5,8]$ such that $f'(c) = fracf(8)-f(-5)8-(-5)$? Your question is not clear.
$endgroup$
– rogerl
Mar 31 at 22:18
$begingroup$
Are you trying to find all $cin [-5,8]$ such that $f'(c) = fracf(8)-f(-5)8-(-5)$? Your question is not clear.
$endgroup$
– rogerl
Mar 31 at 22:18
$begingroup$
You evaluated $f'(b)$, not $f(b)$ (and same for $a$)
$endgroup$
– J. W. Tanner
Mar 31 at 22:20
$begingroup$
You evaluated $f'(b)$, not $f(b)$ (and same for $a$)
$endgroup$
– J. W. Tanner
Mar 31 at 22:20
$begingroup$
@rogerl I don't understand your math symbol $epsilon$ (I only know it's name). I have applied the Mean Value Theorem to $f(x) = 2x^3-6x^2-90x+6$ to find out $f'(c) = frac-13013$ now I need to find all values of $c$ such that $f'(c) = frac-13013$
$endgroup$
– LuminousNutria
Mar 31 at 22:21
$begingroup$
@rogerl I don't understand your math symbol $epsilon$ (I only know it's name). I have applied the Mean Value Theorem to $f(x) = 2x^3-6x^2-90x+6$ to find out $f'(c) = frac-13013$ now I need to find all values of $c$ such that $f'(c) = frac-13013$
$endgroup$
– LuminousNutria
Mar 31 at 22:21
1
1
$begingroup$
$in$ means "is contained in" (the interval)
$endgroup$
– J. W. Tanner
Mar 31 at 22:22
$begingroup$
$in$ means "is contained in" (the interval)
$endgroup$
– J. W. Tanner
Mar 31 at 22:22
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You are overcomplicating things; there is no need to apply the mean value theorem to $f'$.
To find all values of $cin[-5,8]$ for which $f'(c)=-10$ you indeed want to first compute $f'$. You correctly found that
$$f'(c)=6c^2-12c-90,$$
and setting this equal to $-10$ yields
$$6c^2-12c-90=-10.$$
This is a quadratic equation in $c$, which you can solve yourself, judging from your attempted solution.
Caution: The quadratic equation above has (at most) two solutions. Make sure to check whether the solutions are in fact in the interval $[-5,8]$.
answered Mar 31 at 22:22
ServaesServaes
30.6k342101
$endgroup$
add a comment |
$begingroup$
You got the wrong answer because you set $f'(c)=fracf'(b)-f'(a)b-a$ instead of $fracf(b)-f(a)b-a.$
Otherwise your approach was correct.
answered Mar 31 at 22:26
J. W. TannerJ. W. Tanner
4,7971420
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
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$begingroup$
You are overcomplicating things; there is no need to apply the mean value theorem to $f'$.
To find all values of $cin[-5,8]$ for which $f'(c)=-10$ you indeed want to first compute $f'$. You correctly found that
$$f'(c)=6c^2-12c-90,$$
and setting this equal to $-10$ yields
$$6c^2-12c-90=-10.$$
This is a quadratic equation in $c$, which you can solve yourself, judging from your attempted solution.
Caution: The quadratic equation above has (at most) two solutions. Make sure to check whether the solutions are in fact in the interval $[-5,8]$.
answered Mar 31 at 22:22
ServaesServaes
30.6k342101
$endgroup$
add a comment |
$begingroup$
You are overcomplicating things; there is no need to apply the mean value theorem to $f'$.
To find all values of $cin[-5,8]$ for which $f'(c)=-10$ you indeed want to first compute $f'$. You correctly found that
$$f'(c)=6c^2-12c-90,$$
and setting this equal to $-10$ yields
$$6c^2-12c-90=-10.$$
This is a quadratic equation in $c$, which you can solve yourself, judging from your attempted solution.
Caution: The quadratic equation above has (at most) two solutions. Make sure to check whether the solutions are in fact in the interval $[-5,8]$.
answered Mar 31 at 22:22
ServaesServaes
30.6k342101
$endgroup$
add a comment |
$begingroup$
You are overcomplicating things; there is no need to apply the mean value theorem to $f'$.
To find all values of $cin[-5,8]$ for which $f'(c)=-10$ you indeed want to first compute $f'$. You correctly found that
$$f'(c)=6c^2-12c-90,$$
and setting this equal to $-10$ yields
$$6c^2-12c-90=-10.$$
This is a quadratic equation in $c$, which you can solve yourself, judging from your attempted solution.
Caution: The quadratic equation above has (at most) two solutions. Make sure to check whether the solutions are in fact in the interval $[-5,8]$.
answered Mar 31 at 22:22
ServaesServaes
30.6k342101
$endgroup$
You are overcomplicating things; there is no need to apply the mean value theorem to $f'$.
To find all values of $cin[-5,8]$ for which $f'(c)=-10$ you indeed want to first compute $f'$. You correctly found that
$$f'(c)=6c^2-12c-90,$$
and setting this equal to $-10$ yields
$$6c^2-12c-90=-10.$$
This is a quadratic equation in $c$, which you can solve yourself, judging from your attempted solution.
Caution: The quadratic equation above has (at most) two solutions. Make sure to check whether the solutions are in fact in the interval $[-5,8]$.
answered Mar 31 at 22:22
ServaesServaes
30.6k342101
answered Mar 31 at 22:22
ServaesServaes
30.6k342101
answered Mar 31 at 22:22
ServaesServaes
30.6k342101
answered Mar 31 at 22:22
ServaesServaes
30.6k342101
30.6k342101
add a comment |
add a comment |
$begingroup$
You got the wrong answer because you set $f'(c)=fracf'(b)-f'(a)b-a$ instead of $fracf(b)-f(a)b-a.$
Otherwise your approach was correct.
answered Mar 31 at 22:26
J. W. TannerJ. W. Tanner
4,7971420
$endgroup$
add a comment |
$begingroup$
You got the wrong answer because you set $f'(c)=fracf'(b)-f'(a)b-a$ instead of $fracf(b)-f(a)b-a.$
Otherwise your approach was correct.
answered Mar 31 at 22:26
J. W. TannerJ. W. Tanner
4,7971420
$endgroup$
add a comment |
$begingroup$
You got the wrong answer because you set $f'(c)=fracf'(b)-f'(a)b-a$ instead of $fracf(b)-f(a)b-a.$
Otherwise your approach was correct.
answered Mar 31 at 22:26
J. W. TannerJ. W. Tanner
4,7971420
$endgroup$
You got the wrong answer because you set $f'(c)=fracf'(b)-f'(a)b-a$ instead of $fracf(b)-f(a)b-a.$
Otherwise your approach was correct.
answered Mar 31 at 22:26
J. W. TannerJ. W. Tanner
4,7971420
answered Mar 31 at 22:26
J. W. TannerJ. W. Tanner
4,7971420
answered Mar 31 at 22:26
J. W. TannerJ. W. Tanner
4,7971420
answered Mar 31 at 22:26
J. W. TannerJ. W. Tanner
4,7971420
4,7971420
add a comment |
add a comment |
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$begingroup$
Are you trying to find all $cin [-5,8]$ such that $f'(c) = fracf(8)-f(-5)8-(-5)$? Your question is not clear.
$endgroup$
– rogerl
Mar 31 at 22:18
$begingroup$
You evaluated $f'(b)$, not $f(b)$ (and same for $a$)
$endgroup$
– J. W. Tanner
Mar 31 at 22:20
$begingroup$
@rogerl I don't understand your math symbol $epsilon$ (I only know it's name). I have applied the Mean Value Theorem to $f(x) = 2x^3-6x^2-90x+6$ to find out $f'(c) = frac-13013$ now I need to find all values of $c$ such that $f'(c) = frac-13013$
$endgroup$
– LuminousNutria
Mar 31 at 22:21
1
$begingroup$
$in$ means "is contained in" (the interval)
$endgroup$
– J. W. Tanner
Mar 31 at 22:22