State the Bolzano-Weierstrass theorem. Use it to show that a continuous function $f:[a, b] \rightarrow \mathbb{R}$ attains a global maximum; that is, there is a real number $c \in[a, b]$ such that $f(c) \geqslant f(x)$ for all $x \in[a, b]$.

A function $f$ is said to attain a local maximum at $c \in \mathbb{R}$ if there is some $\varepsilon>0$ such that $f(c) \geqslant f(x)$ whenever $|x-c|<\varepsilon$. Suppose that $f: \mathbb{R} \rightarrow \mathbb{R}$ is twice differentiable, and that $f^{\prime \prime}(x)<0$ for all $x \in \mathbb{R}$. Show that there is at most one $c \in \mathbb{R}$ at which $f$ attains a local maximum.

If there is a constant $K<0$ such that $f^{\prime \prime}(x)<K$ for all $x \in \mathbb{R}$, show that $f$ attains a global maximum. [Hint: if $g^{\prime}(x)<0$ for all $x \in \mathbb{R}$, then $g$ is decreasing.]

Must $f: \mathbb{R} \rightarrow \mathbb{R}$ attain a global maximum if we merely require $f^{\prime \prime}(x)<0$ for all $x \in \mathbb{R} ?$ Justify your answer.