Let g(x) = f(x) + f(1 - x) and f''(x)

Question

(D) Given $g(x) = f(x) + f(1 - x)$.Differentiating with respect to $x$, we get $g'(x) = f'(x) - f'(1 - x)$.Since $f''(x) < 0$, $f'(x)$ is a strictly d

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$g(x)$ decreases on $[1/2, 1]$ and $g(x)$ increases on $[0, 1/2]$

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(D) Given $g(x) = f(x) + f(1 - x)$.Differentiating with respect to $x$, we get $g'(x) = f'(x) - f'(1 - x)$.Since $f''(x) For $x \in [0, 1/2]$, we have $x \le 1 - x$, which implies $f'(x) \ge f'(1 - x)$ because $f'(x)$ is decreasing.Thus, $g'(x) = f'(x) - f'(1 - x) \ge 0$ for $x \in [0, 1/2]$, so $g(x)$ is increasing on $[0, 1/2]$.For $x \in [1/2, 1]$, we have $x \ge 1 - x$, which implies $f'(x) \le f'(1 - x)$ because $f'(x)$ is decreasing.Thus, $g'(x) = f'(x) - f'(1 - x) \le 0$ for $x \in [1/2, 1]$, so $g(x)$ is decreasing on $[1/2, 1]$.Therefore, $g(x)$ increases on $[0, 1/2]$ and decreases on $[1/2, 1]$.

Let $g(x) = f(x) + f(1 - x)$ and $f''(x) < 0$ for $0 \le x \le 1$. Which of the following is true regarding the monotonicity of $g(x)$?

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