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 $ ?

If there were three distinct roots $x_1<x_2<x_3$ of $f(x)-g(x)=0$ then by choosing $\lambda_2$ so that $x_2=\lambda_2x_1+(1-\lambda_2)x_3$ we would have
$$
(f-g)(x_2)=0=\lambda_2(f-g)(x_1)+(1-\lambda_2)(f-g)(x_3),
$$
which would be in contradiction with the strict convexity of $f-g$.
