Prove a convex and concave function can have at most 2 solutions Let $ a,b \in \mathbb{R}$
Let $f(x) : [a,b] \rightarrow \mathbb{R} $ is a continuous function in $ [a ,b] $, differentiable and strictly convex in $ (a, b) $
and $g(x) : [a,b] \rightarrow \mathbb{R} $ is a continuous function in $ [a ,b] $, differentiable and strictly concave in $ (a, b) $
How can I prove the intersection of $ f $ and $ g $ can have a maximum of two roots $f(x)-g(x)=0 $ ?

$h(x)=f(x)-g(x)$ is a strictly convex function in $[a,b]$.
So we just have to prove that a convex function has at most $2$ roots in $[a,b]$.
If $h(x)$ has $3$ roots let's say, $p,q,r$ then there is $c \in (p,q)$ such that $h'(c)=0$ (by Rolle's Theorem) and there is $d \in (q,r)$ such that $h'(d)=0$.
But it means that there is $e \in (c,d)$ such that $(h')'(e)=0$, which is a contradiction because $h''(x)>0$. 
