Prove using Rolle's theorem that $x^2$ has $1$ root I'm having a hard time using Rolle's theorem in this proof and my professor said that it will likely be an exam question similar to the one above.
Let $f(x) = x^2$. Now $f(0) = 0$, and $f$ is strictly increasing on $(0, +\infty)$.
Suppose there is another $a &gt; 0$ such that $f(0) = f(a) = 0$. Then by Rolle's theorem there exists a point $c \in (0, a)$ such that
$$f'(c) = 0,$$
but this is clearly impossible, because as we said earlier $f$ is strictly increasing and therefore $f'(x) &gt; 0$ on $(0, +\infty)$. The same argument can be applied to the interval $(-\infty, 0)$ and we conclude that $0$ is the only root.
Hint: Consider the function $x^2$ over the interval $[-a,a]$ 


  Rolle's theorem: If a real-valued function $f$ is continuous on a closed interval $[a, b]$, differentiable on the open interval $(a, b)$, and $f(a) = f(b)$, then there exists at least one $c$ in the open interval $(a, b)$ such that
  $$f'(c) = 0.$$