Question

Let $g : \mathbb{R} \rightarrow \mathbb{R}$ be a non-constant twice differentiable function such that $g'\left(\frac{1}{2}\right) = g'\left(\frac{3}{2}\right)$. If a real-valued function $f$ is defined as $f(x) = \frac{1}{2} \left[ g(x) + g(2 - x) \right],$ then

• A. $f''(x) = 0$ for at least two $x$ in $(0, 2)$
• B. $f''(x) = 0$ for exactly one $x$ in $(0, 1)$
• C. $f''(x) = 0$ for no $x$ in $(0, 1)$
• D. $f'\left(\frac{3}{2}\right) + f'\left(\frac{1}{2}\right) = 1$

Answer 1

Answer: (A)

$f''(x) = 0$ for at least two $x$ in $(0, 2)$

Answer 2

Since $f(x) = \frac{1}{2} [ g(x) + g(2 - x) ]$, we observe that $f(x)$ is symmetric about $x = 1$, suggesting that the behavior around $x = 1$ is crucial.

Calculate $f'(x)$:
$f'(x) = \frac{1}{2} \left[ g'(x) + g'(2 - x) \right]$

Given $g'\left( \frac{1}{2} \right) = g'\left( \frac{3}{2} \right)$, we find:
$f'\left( \frac{1}{2} \right) = \frac{1}{2} \left[ g'\left( \frac{1}{2} \right) + g'\left( \frac{3}{2} \right) \right] = 0$
and similarly,
$f'\left( \frac{3}{2} \right) = 0$

Calculate $f''(x)$:
$f''(x) = \frac{1}{2} \left[ g''(x) - g''(2 - x) \right]$

Since $g$ is non-constant and twice differentiable, by the Intermediate Value Theorem, $f''(x) = 0$ must occur at least twice in $(0, 2)$.