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Suppose that $f:[0,\to \infty) \to \mathbf{R}$ is a continuous function such that $f(0)=0$ and $f(x)\ge \sqrt{x}$ for all $x\ge 0$. Show that for each $r>0$ there exist $c>0$ such that $f(c)=r$.

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    Let $r>0$. If there is no $c>0$ such that $f(c)=r$, then show that $f(x)$x$ (use IVT). So this contradicts the assumtion. – 2017-02-15

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$r^2 > 0$ so $f(r^2) > \sqrt{r^2} = r > 0 = f(0)$.

So intermediate value theorem states there for any $y; f(0) < y < f(r^2)$ there is a $c_y; 0 < c_y < r^2$ so that $f(c_y) = y$. Let $y = r$ and we are done.