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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(c)=f(b)f(a)baf^{\prime}(c)=\frac{f(b)-f(a)}{b-a}. If the Mean Value Theorem cannot be applied, explain why not. f(x)=x6f(x)=x^{6}, [-1, 1]

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Answered 9 months ago
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f(x)=x6,x[1,1]\begin{equation*} f(x) = x^6, \hspace{1cm} x \in [-1, 1] \end{equation*}

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

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

We start by finding the derivation of ff.

f(x)=6x5\begin{equation*} f'(x) = 6x^5 \end{equation*}

Now lets evaluate ff at its endpoints

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

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

we have

f(c)=6c5=111+1=f(b)f(a)ba6c5=0c=0\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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