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**SlipEternal** If $h(x) = g(x)+g(2-x)$ then $h'(x) = g'(x)+g'(2-x)\dfrac{d}{dx}\left(2-x\right) = g'(x) -g'(2-x)$. Plugging in $x=1$ gives $h'(1) = g'(1)-g'(2-1) = g'(1)-g'(1) = 0$.

If you have not yet learned the chain rule, then this result is not meaningful to you. In that case, the book is using the Mean Value Theorem instead of the Chain Rule.

$$h(2)-h(0) = g(2)+g(2-2)-(g(0)+g(2-0)) = g(2)+g(0)-g(0)-g(2) = 0$$

Also, by the Mean Value Theorem:

$$h(2)-h(0) = ch'(c)$$

for some value of $c \in (0,2)$. Therefore, there exists $c \in (0,2)$ such that $ch'(c) = 0$. Either $c=0$ or $h'(c)=0$. But, $c \in (0,2)$ means that $c\neq 0$. Therefore, $h'(c) = 0$.