# Solutions of $f(s)=s^2$ with $s\ge 2$ given some conditions on a differentiable function $f$

A differentiable function $$f :\mathbb R \to\mathbb R$$ satisfies:

1. $$f(2) = 5$$
2. $$f(3) = 8$$
3. $$f'(x) \le 4 \sin^2(\pi x)$$ for all $$x \in\mathbb R$$.

How many values of $$s \ge 2$$ satisfy $$f(s) = s^2$$ ?

I could conclude that $$x < 3$$ by the 3rd condition, and tried to use the Mean Value Theorem but found it hard to go on. Can anyone give some hints?

Thanks

Let $g(x)=x^2$ and define a new function $h(x)=g(x)-f(x)$. The question asks for the number of solutions to $f(x)=g(x)$ or equivalently $h(x)=g(x)-f(x)=0$ where $x\ge2$.

All we have to do is note that when $x>2$, we have $$h'(x)=(g(x)-f(x))'=g'(x)-f'(x)=2x-f'(x)\ge 2x-4\sin^2(\pi x)\ge 2x-4>0$$

Now since $h(2)=g(2)-f(2)=-1<0$ while $h(3)=g(3)-f(3)=1>0$ we know by the intermediate value theorem that there is at least one solution to $h(x)=0$ and since $h'(x)>0$ for $x>2$, we know this solution is unique.