Continuous function and extreme value theorem. Suppose $f(x)$ is a continued function defined on $[-1,1]$ with $f(0)&gt;0$ and $f(\pm 1)=0$. Prove that there exist constants $a&gt;0$ and $b$ such that $g(x)=-a\vert x\vert +b\geq f(x) (-1\leq x \leq 1)$, and for which there exists $c\in (-1,1)$ such that $g(c)=f(c)$.

I don't have a clue about how to solve this problem. The book gives a hint that tells me to use the extreme value theorem, and I don't know how to make use of it. And it seems that if I want to use contradiction I would have to construct a very complicated negation.
Pick $a$ s.t. $0 &lt; a &lt; f(0)$ (for example $\dfrac{f(0)}{2}$ will do).

Let $h(x) = a\vert x\vert+ f(x)$. Since $h$ is continuous on $[-1,1]$ and $h(1) = h(-1) = a &lt; h(0) = f(0)$, it has a maximum $c$ strictly inside the interval, so $c$ is in $(-1,1)$.

Then if $b = h(c)$, you get your $g(x) = -a\vert x\vert + b, g(c) = f(c)$ and $g(x) \geq f(x)$ since $f(x) - g(x) = h(x) - h(c)$ and $c$ is the maximum of $h$ by definition.
Because $[-1,1]$ is compact, $f$ has a max, suppose k and note that $k&gt;0$ because exist a point (0) where f is positive. If we put $b=k+1$ and $a=1$ is easy check that works. The second item is not necessary true...