<s>[INST] Applying the Mean Value Theorem to conclude a function has a zero 
Consider the function $f$ given by $f(x)=(x-2)^4\cos(x^2-4x+4)$. Use the Mean Value Theorem to show that $f'$ has a zero on the interval on $[1,3]$.

I notice that to do this we must show $f'(c)=0$ where $c$ is real number in the interval $[1,3]$. Now by the Mean Value Theorem,
$$\frac{f(3)-f(1)}{3-1} =f'(c)\,.$$
Notice that $f'(c)$ is indeed $0$ on the left hand side.
 [/INST] First note that $x^2-4x+4 = (x-2)^2.$
Then the left-hand side of your MVT equation gives
\begin{align}
\frac{f(3) - f(1)}{3-1} &= \frac{(3-2)^4 \cos (3-2)^2 - (1-2)^4 \cos (1-2)^2}{2} \\
&= \frac{(1)^4 \cos (1)^2 - (-1)^4 \cos (-1)^2}{2} \\
&= \frac{\cos(1) - \cos(1)}{2} \\
&= 0,
\end{align}
which equals $f'(c)$ for some $c \in (1,3).$
(Notice that my interval is open.  I think this is what you intended to type.)
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