Find all numbers $c$ that satisfy the conclusion of the Mean Value Theorem for the following function and interval: $$f(x)=9x^3+9x-7$$ and $[0,2]$.
As far as I know I take the derivative of the function and set it equal to $0$ and $2$, but that isn't working. When I did that I got $\pm\sqrt{\frac{-1}{3}}$ or $\pm \sqrt{\frac{-7}{27}}$.
Best Answer
We have that: $$f(0) = -7$$ $$f(2) = 83$$ $$f'(x) = 27x^2 + 9$$
Then, there exists a $c in \ (0,2)$ such that:
$$f'(c) = \frac{f(2) - f(0)}{2- 0 } = \frac{83 + 7}{2} = 45$$
This means that you have to solve the following:
$$f'(c) = 45 \Rightarrow 27c^2 + 9 = 45 \Rightarrow 3c^2 - 4 = 0$$
Solutions are $c = \pm \frac{2\sqrt{3}}{3} = \pm 1.1547\ldots$
Only $c = \frac{2\sqrt{3}}{3}$ is inside the set $(0,2)$ and hence this $c$ is what you are looking for.