Explain what it means for a bounded function $f:[a, b] \rightarrow \mathbb{R}$ to be Riemann integrable.

Let $f:[0, \infty) \rightarrow \mathbb{R}$ be a strictly decreasing continuous function. Show that for each $x \in(0, \infty)$, there exists a unique point $g(x) \in(0, x)$ such that

$$\frac{1}{x} \int_{0}^{x} f(t) d t=f(g(x)) .$$

Find $g(x)$ if $f(x)=e^{-x}$.

Suppose now that $f$ is differentiable and $f^{\prime}(x)<0$ for all $x \in(0, \infty)$. Prove that $g$ is differentiable at all $x \in(0, \infty)$ and $g^{\prime}(x)>0$ for all $x \in(0, \infty)$, stating clearly any results on the inverse of $f$ you use.