Let f: [0, 2] to R be a twice differentiable | vedclass

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(C) Given $\phi(x) = f(x) + f(2 - x)$.Differentiating with respect to $x$,we get $\phi'(x) = f'(x) - f'(2 - x)$.Since $f''(x) > 0$ for all $x \in (0,

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decreasing on $(0, 1)$ and increasing on $(1, 2)$

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(C) Given $\phi(x) = f(x) + f(2 - x)$.Differentiating with respect to $x$,we get $\phi'(x) = f'(x) - f'(2 - x)$.Since $f''(x) > 0$ for all $x \in (0, 2)$,the derivative $f'(x)$ is a strictly increasing function on $(0, 2)$.Case $I$: If $x > 1$,then $x > 2 - x$. Since $f'(x)$ is strictly increasing,$f'(x) > f'(2 - x)$,which implies $\phi'(x) = f'(x) - f'(2 - x) > 0$. Thus,$\phi(x)$ is increasing on $(1, 2)$.Case $II$: If $x Therefore,$\phi(x)$ is decreasing on $(0, 1)$ and increasing on $(1, 2)$.

Let $f: [0, 2] \to R$ be a twice differentiable function such that $f''(x) > 0$ for all $x \in (0, 2)$. If $\phi(x) = f(x) + f(2 - x)$,then $\phi$ is

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