Let phi(x) = f(x) + f(1-x) and f^{prime prime | vedclass

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(A) Given $\phi(x) = f(x) + f(1-x)$.To determine the monotonicity, we find the derivative $\phi^{\prime}(x) = f^{\prime}(x) - f^{\prime}(1-x)$.Since $

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$\phi$ is monotonic increasing in $\left[0, \frac{1}{2}\right]$ and monotonic decreasing in $\left[\frac{1}{2}, 1\right]$

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(A) Given $\phi(x) = f(x) + f(1-x)$.To determine the monotonicity, we find the derivative $\phi^{\prime}(x) = f^{\prime}(x) - f^{\prime}(1-x)$.Since $f^{\prime \prime}(x) If $x  f^{\prime}(1-x)$ because $f^{\prime}$ is decreasing.Thus, $\phi^{\prime}(x) = f^{\prime}(x) - f^{\prime}(1-x) > 0$ for $x \in \left[0, \frac{1}{2}\right]$, so $\phi(x)$ is increasing in $\left[0, \frac{1}{2}\right]$.If $x > \frac{1}{2}$, then $x > 1-x$, which implies $f^{\prime}(x) Thus, $\phi^{\prime}(x) = f^{\prime}(x) - f^{\prime}(1-x) < 0$ for $x \in \left[\frac{1}{2}, 1\right]$, so $\phi(x)$ is decreasing in $\left[\frac{1}{2}, 1\right]$.

Let $\phi(x) = f(x) + f(1-x)$ and $f^{\prime \prime}(x) < 0$ in $[0, 1]$, then

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