If f: R rightarrow R is an even function whic | vedclass

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(C) Given that $f(x)$ is an even function,by definition,$f(x) = f(-x)$ for all $x \in R$. Differentiating both sides with respect to $x$,we get $f^{\p

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$1$

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(C) Given that $f(x)$ is an even function,by definition,$f(x) = f(-x)$ for all $x \in R$. Differentiating both sides with respect to $x$,we get $f^{\prime}(x) = -f^{\prime}(-x)$. Differentiating again with respect to $x$,we get $f^{\prime \prime}(x) = f^{\prime \prime}(-x)$. This shows that the second derivative of an even function is also an even function. Since $f^{\prime \prime}(x)$ is an even function,$f^{\prime \prime}(-\pi) = f^{\prime \prime}(\pi)$. Given $f^{\prime \prime}(\pi) = 1$,it follows that $f^{\prime \prime}(-\pi) = 1$.

If $f: R \rightarrow R$ is an even function which is twice differentiable on $R$ and $f^{\prime \prime}(\pi)=1$,then $f^{\prime \prime}(-\pi)$ is equal to

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