MA005: Calculus I

Graphically, a function is concave up if its graph is curved with the opening upward (Fig. 1a). Similarly, a function is concave down if its graph opens downward (Fig. 1b). The concavity of a function can be important in applied problems and can even affect billion-dollar decisions.

Fig. 1

An Epidemic: Suppose an epidemic has started, and you, as a member of congress, must decide whether the current methods are effectively fighting the spread of the disease or whether more drastic measures and more money are needed. In Fig. 2, $\mathrm{f}(\mathrm{x})$ is the number of people who have the disease at time $\mathrm{x}$, and two different situations are shown. In both $(a)$ and $(b)$, the number of people with the disease, $f$ now), and the rate at which new people are getting sick, $f'$ (now), are the same. The difference in the two situations is the concavity of $\mathrm{f}$, and that difference in concavity might have a big effect on your decision. In $(a)$, $f$ is concave down at "now", and it appears that the current methods are starting to bring the epidemic under control. In $(b)$, $\mathrm{f}$ is concave up, and it appears that the epidemic is still out of control.

Fig. 2

Usually it is easy to determine the concavity of a function by examining its graph, but we also need a definition which does not require that we have a graph of the function, a definition we can apply to a function described by a formula without having to graph the function.

Fig. 3 shows the concavity of a function at several points. The next theorem gives an easily applied test for the concavity of a function given by a formula.

Fig. 3

Proof: (a) Assume that $\mathrm{f}^{\prime \prime}(\mathrm{x}) > 0$ for all $\mathrm{x}$ in $\mathrm{I}$, and let a be any point in $\mathrm{I}$. We want to show that $\mathrm{f}$ is concave up at $a$ so we need to prove that the graph of $f$ (Fig. 4) is above the tangent line to $\mathrm{f}$ at $a$: $\mathrm{f}(\mathrm{x}) > \mathrm{L}(\mathrm{x})=\mathrm{f}(\mathrm{a})+\mathrm{f}^{\prime}(\mathrm{a})(\mathrm{x}-\mathrm{a})$ for $\mathrm{x}$ close to $a$.

Fig. 4

Assume that $\mathrm{x}$ is in $\mathrm{I}$, and apply the Mean Value Theorem to $\mathrm{f}$ on the interval from $a$ to $\mathrm{x}$. Then there is a number $\mathrm{c}$ between $a$ and $\mathrm{x}$ so that $\mathrm{f}^{\prime}(\mathrm{c})=\frac{\mathrm{f}(\mathrm{x})-\mathrm{f}(\mathrm{a})}{\mathrm{x}-\mathrm{a}} \text { and } \mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{a})+\mathrm{f}^{\prime}(\mathrm{c})(\mathrm{x}-\mathrm{a})$

Since $\mathrm{f} " > 0$ between $a$ and $\mathrm{x}$, we know from the Second Shape Theorem that $\mathrm{f}^{\prime}$ is increasing between $a$ and $\mathrm{x}$. We will consider two cases: $\mathrm{x} > \mathrm{a}$ and $\mathrm{x} < \mathrm{a}$.

If $x > a$, then $x-a > 0$ and $c$ is in the interval $[a, x]$ so $a < c$. Since $f^{\prime}$ is increasing, $a < c$ implies that $\mathrm{f}^{\prime}(\mathrm{a}) < \mathrm{f}^{\prime}(\mathrm{c})$. Multiplying each side of the inequality $\mathrm{f}^{\prime}(\mathrm{a}) < \mathrm{f}^{\prime}(\mathrm{c})$ by the positive amount $\mathrm{x}-\mathrm{a}$, we get that $\mathrm{f}^{\prime}(\mathrm{a})(\mathrm{x}-\mathrm{a}) < \mathrm{f}^{\prime}(\mathrm{c})(\mathrm{x}-\mathrm{a})$. Adding $\mathrm{f}(\mathrm{a})$ to each side of this last inequality, we have $\mathrm{L}(\mathrm{x}) = \mathrm{f}(\mathrm{a})+\mathrm{f}^{\prime}(\mathrm{a})(\mathrm{x}-\mathrm{a}) < \mathrm{f}(\mathrm{a})+\mathrm{f}^{\prime}(\mathrm{c})(\mathrm{x}-\mathrm{a})=\mathrm{f}(\mathrm{x})$.

If $x < a$, then $x-a < 0$ and $c$ is in the interval $[x, a]$ so $c < a$. Since $f^{\prime}$ is increasing, $c <$ a implies that $\mathrm{f}^{\prime}(\mathrm{c}) < \mathrm{f}^{\prime}(\mathrm{a})$. Multiplying each side of the inequality $\mathrm{f}^{\prime}(\mathrm{c}) < \mathrm{f}^{\prime}(\mathrm{a})$ by the negative amount $\mathrm{x}-\mathrm{a}$, we get that $\mathrm{f}^{\prime}(\mathrm{c})(\mathrm{x}-\mathrm{a})>\mathrm{f}^{\prime}(\mathrm{a})(\mathrm{x}-\mathrm{a})$ and $\mathrm{f}(\mathrm{x})=\mathrm{f}(\mathrm{a})+\mathrm{f}^{\prime}(\mathrm{c})(\mathrm{x}-\mathrm{a}) > \mathrm{f}(\mathrm{a})+\mathrm{f}^{\prime}(\mathrm{a})(\mathrm{x}-\mathrm{a})=\mathrm{L}(\mathrm{x})$.

In each case we get that the function $\mathrm{f}(\mathrm{x})$ is above the tangent line $\mathrm{L}(\mathrm{x})$. The proof of $(b)$ is similar.

(c) Let $(x)=x^{4}, g(x)=-x^{4}$, and $h(x)=x^{3}$ (Fig.5). The second derivative of each of these functions is zero at $a=0$, and at $(0,0)$ they all have the same tangent line: $L(x)=0$, the $x$-axis. However, at $(0,0)$ they all have different concavity: $\mathrm{f}$ is concave up, $g$ is concave down, and $\mathrm{h}$ is neither concave up nor down.

Fig. 5

Practice 1: Use the graph of $\mathrm{f}$ in Fig. 6 to finish filling in the table with "$+$" for positive, "$-$" for negative" $,$ or "$0$".

$\begin{array}{l|c|c|c|c}\mathrm{x} & \mathrm{f}(\mathrm{x}) & \mathrm{f}^{\prime}(\mathrm{x}) & \mathrm{f} "(\mathrm{x}) & \text { Concavity (up or down) } \\\hline 1 & + & + & - & \text { down } \\2 & + & & & \\3 & - & & & \\4 & & & &\end{array}$

Fig. 6