<s>[INST] 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 $[-5, 8]$ 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 $f'(c) = \frac{-130}{13}=-10$.
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 $f'(c) = \frac{f(b)-f(a)}{b - a}$ again. 
$f(b) = 6(-5)^2-12(-5)-90 = 150+60-90 = 120$
$f(a) = 6(8)^2-12(8)-90 = 384-186 = 198$
$f'(c) = \frac{120-198}{-5-8} = \frac{78}{13} = 6$ 
Then I took $6x^2-12x-90$ again, replaced $x$ with $c$, and set it equal to $6$ before solving for $c$.
$6c^2-12c-90 = 6  \to 6(c^2-2c-16) = 0$
$\frac{2 \pm \sqrt{2^2-4(1)(-17)}}{2(1)} = \frac{2 \pm \sqrt{74}}{2}$ or $c = \frac{2 - \sqrt{74}}{2}, \frac{2 + \sqrt{74}}{2}$
But this answer is wrong.
How can I find the values for $c$ here?
 [/INST] You are overcomplicating things; there is no need to apply the mean value theorem to $f'$.
To find all values of $c\in[-5,8]$ 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 $[-5,8]$.
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