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# Determine whether the Mean Value Theorem can be applied to f on the closed interval [a, b]. If the Mean Value Theorem can be applied, find all values of c in the open interval (a, b) such that $f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$. If the Mean Value Theorem cannot be applied, explain why not. $f(x)=x^{6}$, [-1, 1]

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$\begin{equation*} f(x) = x^6, \hspace{1cm} x \in [-1, 1] \end{equation*}$

The mean value theorem tells us that if $f$ is a continuous function on a closed interval $[a,b]$, differentiable on the open interval $(a,b)$ then there is a number $c \in [a,b]$ so that $f'(c) = \dfrac{f(b) - f(a)}{b-a}$.

As our function is continuous and differentiable on $[-1, 1]$ we can apply the mean value theorem.

We start by finding the derivation of $f$.

$\begin{equation*} f'(x) = 6x^5 \end{equation*}$

Now lets evaluate $f$ at its endpoints

\begin{align*} f(-1) = 1 \\ f(1) = 1 \end{align*}

We have to find $c \in [-1, 1]$ so that $f'(c) = \dfrac{f(b) - f(a)}{b-a}$.

we have

\begin{align*} f'(c) = 6c^5 &= \dfrac{1 - 1}{1 + 1} = \dfrac{f(b) - f(a)}{b-a} \\ \\ 6c^5 &= 0 \\ & \Rightarrow c = 0 \end{align*}

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