If f''(x)

Question

(B) Given $f''(x) < 0$,which implies that $f'(x)$ is a strictly decreasing function. Let $H(x) = f(1 - x) + 2f(x/2)$. Differentiating with respect to

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decreasing in $(2/3, 2)$ and increasing in $(0, 2/3)$

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(B) Given $f''(x) Let $H(x) = f(1 - x) + 2f(x/2)$. Differentiating with respect to $x$,we get $H'(x) = -f'(1 - x) + f'(x/2)$. For $H(x)$ to be increasing,we require $H'(x) > 0$,which means $-f'(1 - x) + f'(x/2) > 0$,or $f'(x/2) > f'(1 - x)$. Since $f'(x)$ is a decreasing function,$f'(a) > f'(b)$ implies $a Therefore,$x/2 Solving for $x$: $x/2 + x Thus,$H(x)$ is increasing in $(0, 2/3)$. Similarly,for $H(x)$ to be decreasing,we require $H'(x)  1 - x$,or $x > 2/3$. Thus,$H(x)$ is decreasing in $(2/3, 2)$.

If $f''(x) < 0$ for all $x \in (0, 2)$,then the function $H(x) = f(1 - x) + 2f(x/2)$ is:

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