Seeing the increasing or decreasing tendency of $h(x)$

Question

$$h(x)=3f(\frac{x^2}{3})+f(3-x^2)$$ for every $x$ in $(-3,4), where\quad $f"(x)>0$ \quad in \quad the\quad domain \quad of \qquad $h(x). Then h(x) is:

$(a)$ increasing in $(\frac{3}{2},4)$ and $(\frac{-3}{2},0)$

$(b)$ decreasing in $(-3,\frac{-3}{2})$ and $(0,\frac{3}{2})$

My attempt:

I can see that $f'(x)$ will be increasing function. But from there I am not able to figure out what will be the nature of $h"(x)$.

Answer 1

For $h(x)=3f(\frac{x^2}{3})+f(3-x^2)$ to be increasing, $$h'(x) = 2xf'(\frac{x^2}{3}) - 2xf'(3 - x^2) = 2x(f'(\frac{x^2}{3}) - f'(3 - x^2)) > 0 $$ in the domain $(-3, 4)$ which happens iff

$x > 0 $ and $\frac{x^2}{3} > 3 - x^2 \implies x \in (\frac{3}{2}, 4)$; or

$x < 0 $ and $\frac{x^2}{3} < 3 - x^2 \implies x \in (\frac{-3}{2}, 0)$; since $f''(x) > 0 \forall$ $x$ in domain

Similarly, you can work out the region in which $h(x)$ is decreasing