Let g: R rightarrow R be a non-constant twice | vedclass

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(A) Given $f(x) = \frac{1}{2}[g(x) + g(2-x)]$. Taking the derivative,$f^{\prime}(x) = \frac{1}{2}[g^{\prime}(x) - g^{\prime}(2-x)]$. We are given $g^{

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$f^{\prime}(x)=0$ for at least two $x$ in $(0,2)$

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(A) Given $f(x) = \frac{1}{2}[g(x) + g(2-x)]$. Taking the derivative,$f^{\prime}(x) = \frac{1}{2}[g^{\prime}(x) - g^{\prime}(2-x)]$. We are given $g^{\prime}\left(\frac{1}{2}\right) = g^{\prime}\left(\frac{3}{2}\right)$. Evaluating $f^{\prime}$ at $x = \frac{1}{2}$: $f^{\prime}\left(\frac{1}{2}\right) = \frac{1}{2}[g^{\prime}\left(\frac{1}{2}\right) - g^{\prime}\left(\frac{3}{2}\right)] = 0$. Evaluating $f^{\prime}$ at $x = \frac{3}{2}$: $f^{\prime}\left(\frac{3}{2}\right) = \frac{1}{2}[g^{\prime}\left(\frac{3}{2}\right) - g^{\prime}\left(\frac{1}{2}\right)] = 0$. Since $f^{\prime}\left(\frac{1}{2}\right) = 0$ and $f^{\prime}\left(\frac{3}{2}\right) = 0$,by Rolle's Theorem applied to $f^{\prime}(x)$ on the interval $\left[\frac{1}{2}, \frac{3}{2}\right]$,there exists at least one $c \in \left(\frac{1}{2}, \frac{3}{2}\right)$ such that $f^{\prime \prime}(c) = 0$. Also,note that $f^{\prime}(1) = \frac{1}{2}[g^{\prime}(1) - g^{\prime}(1)] = 0$. Thus,$f^{\prime}(x)$ is zero at $x = \frac{1}{2}, 1, \frac{3}{2}$. By Rolle's Theorem,$f^{\prime \prime}(x)$ must be zero at least once in $\left(\frac{1}{2}, 1\right)$ and at least once in $\left(1, \frac{3}{2}\right)$. Therefore,$f^{\prime}(x) = 0$ for at least three values in $(0, 2)$,which satisfies option $A$.

Let $g: R \rightarrow R$ be a non-constant twice differentiable function such that $g^{\prime}\left(\frac{1}{2}\right)=g^{\prime}\left(\frac{3}{2}\right)$. If a real-valued function $f$ is defined as $f(x)=\frac{1}{2}[g(x)+g(2-x)]$,then:

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