Question

Let $g(x) = \displaystyle \int_{1-x}^{1+x} t\,\bigl|f'(t)\bigr|\,dt$ where $f(x)$ does not behave like a constant function in any interval $(a,b)$ and the graph of $y = f'(x)$ is symmetrical about the line $x = 1$. Then

• A. $g(x)$ is increasing $\forall x \in \mathbb{R}$
• B. $g(x)$ is increasing only for $x > 1$
• C. $g'(x)$ is an even function
• D. $g'(x)$ is symmetrical about the line $x = 2$

Answer 1

Answer: (A), (C)

$g(x)$ is increasing ∀ $x\in\mathbb{R}$; $g'(x)$ is an even function

Answer 2

Compute the derivative by Leibniz rule:

$$g'(x)= (1+x)\bigl|f'(1+x)\bigr| - \bigl(1 - x\bigr)\bigl|f'(1-x)\bigr|\;(-1) = (1+x)\bigl|f'(1+x)\bigr| + (1-x)\bigl|f'(1-x)\bigr|.$$

Since $f'(x)$ is symmetric about $x=1$, we have $\bigl|f'(1+x)\bigr|=\bigl|f'(1-x)\bigr|$. Hence

$$g'(x) = \bigl|f'(1+x)\bigr|\bigl[(1+x)+(1-x)\bigr] = 2\bigl|f'(1+x)\bigr|\;\ge0.$$

• Monotonicity: $g'(x)\ge0$ for all real $x$, so $g$ is increasing on $\mathbb{R}$.
• Evenness:

$$g'(-x)=2\bigl|f'(1-x)\bigr|=2\bigl|f'(1+x)\bigr|=g'(x),$$

so $g'(x)$ is an even function.