Calculus application of derivatives Let $g(x) = 2f(x/2) + f(2-x)$ and $f''(x)<0$ for all $x\in (0,2)$. If $g(x)$ increases in $(a,b)$ and decreases in $(c,d)$ find the values of $a$, $b$, $c$ and $d$.
What I thought was a little graphical approach. I figured it out that if $f''(x)$ is less than zero then $f'(x)$ would be decreasing (not considering the concavity) and the behaviour of $g(x)$ would depend upon $f(x)$. But I couldn't figure out the exact intervals. 
 A: 
I figured it out that if $f''(x)$ is less than zero then $f'(x)$ would be decreasing

It's enough. Look:
$$
g'(x) = 2f'(x/2)\cdot \frac12 + f(2-x)\cdot(-1) = f'(x/2) - f(2-x).
$$
I. $0< x/2< 2-x$; since $f'(x)$ is decreasing then $g'(x)>0$ and $g(x)$ is increasing.
II. $x/2 > 2-x > 0$; since $f'(x)$ is decreasing then $g'(x)<0$ and $g(x)$ is decreasing.
I think you can find $a$, $b$, $c$, $d$ now.
A: $$\begin{align}
g(x)&=2f\left(\frac x2\right)+f(2-x) \\[2 ex]
g'(x)&=f'\left(\frac x2\right)-f'(2-x) \\[2 ex]
g'\left(\frac 43\right)&=f'\left(\frac 23\right)-f'\left(\frac 23\right) \\[2 ex]
 &= 0
\end{align}$$
so $\dfrac 43$ is a critical point for $g(x)$. (I found that critical point by solving $\dfrac x2=2-x$.) $f''(x)<0$ implies that $f'(x)$ is decreasing, so 
$$u<\frac 23 \implies f'(u)>f'\left(\frac 23\right)$$
$$u>\frac 23 \implies f'(u)<f'\left(\frac 23\right)$$
Thus, for $x<\dfrac 43$, $\dfrac x2<\dfrac 23$ and $2-x>\dfrac 23$, so
$$\begin{align}
g'(x)&=f'\left(\frac x2\right)-f'(2-x) \\[2 ex]
 &> f'\left(\frac 23\right)-f'\left(\frac 23\right) \\[2 ex]
 &=0
\end{align}$$
and similarly $x>\dfrac 43\implies g'(x)<0$. Thus $g(x)$ increases for $x<\dfrac 43$ and increases for $x>\dfrac 43$.
Now we know the behavior of $f$ only for $0<x<2$. The function $g$ is then known only for the intersection of $0<\dfrac x2<2$ and $0<2-x<2$, which is again $0<x<2$. Therefore $g(x)$ increases in $\left(0,\dfrac 43\right)$ and decreases in $\left(\dfrac 43,2\right)$ and we don't know what happens outside the interval $(0,2)$.

So $a=0,b=c=\dfrac 43,d=2$.

