Minimum number of solutions for $f=g$

Let $f, g : [−1, 1]rightarrow mathbb{R}$ be odd functions whose derivatives are continuous. You
are given that $|g(x)| < 1$ for all $x in [−1, 1], f(−1) = −1, f(1) = 1$ and $f'(0)<g'(0)$. Then find the minimum possible number of solutions to the equation
$f(x) = g(x)$ in the interval $[−1, 1]$.

My answer is $3$. Because $0$ is a trivial solution and in a neighborhood of $0$ the function $g(x)-f(x)$ is increasing. Now applying Intermediate value property we get another two solutions.