Let g(x) = 2f(frac{x}{2}) + f(2 - x) and f''( | vedclass

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(D) Given $g(x) = 2f(\frac{x}{2}) + f(2 - x)$.Taking the derivative with respect to $x$,we get $g'(x) = 2f'(\frac{x}{2}) \cdot \frac{1}{2} + f'(2 - x)

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$(0, 4/3)$

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(D) Given $g(x) = 2f(\frac{x}{2}) + f(2 - x)$.Taking the derivative with respect to $x$,we get $g'(x) = 2f'(\frac{x}{2}) \cdot \frac{1}{2} + f'(2 - x) \cdot (-1) = f'(\frac{x}{2}) - f'(2 - x)$.We are given $f''(x) For $g(x)$ to be increasing,we require $g'(x) > 0$,which means $f'(\frac{x}{2}) - f'(2 - x) > 0$,or $f'(\frac{x}{2}) > f'(2 - x)$.Since $f'(x)$ is strictly decreasing,$f'(a) > f'(b)$ implies $a Therefore,$\frac{x}{2} Solving for $x$: $\frac{x}{2} + x Thus,$g(x)$ increases in the interval $(0, 4/3)$.

Let $g(x) = 2f(\frac{x}{2}) + f(2 - x)$ and $f''(x) < 0$ for all $x \in (0, 2)$. Then $g(x)$ increases in

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