Diffcal

Differentiating using the chain rule: $g'(x)=f'(x)-f'(1-x)$ When $f''(x)>0$ , $f'(x)$ is increasing. This means that if $a>b$ , then $f'(a)>f'(b)$ for $a,b$ on the interval $(0,1)$ .

Now consider a value x on the interval $(0,1)$ (meaning 1-x is also in this interval): $\begin{aligned}\text{when }x&>\frac{1}{2}:&\text{when }x&<\frac{1}{2}:\\ x&>1-x&x&<1-x\\ f'(x)&>f'(1-x)&f'(x)&<f'(1-x)\\ f'(x)-f'(1-x)&>0&f'(x)-f'(1-x)&<0\\ g'(x)&>0&g'(x)&<0\\ \text{So }g(x)&\text{ is increasing on }\left(\frac{1}{2},1\right)&\text{So }g(x)&\text{ is decreasing on }\left(0,\frac{1}{2}\right)\end{aligned}$