Short and Sweet Calculus

## 4.4 Concavity and Points of Inflection

### 4.4.1 Concavity

Although the sign of the derivative provides information about whether a function is an increasing or decreasing, it does not tell us which way the graph of the function is bending. The graphs of two increasing functions are shown in Figure [fig:Ch6-Concavity-Fig1]. The graph on the left bends upward as the point that traces it moves from left to right. In this case, we say the function is concave upward. The graph on the right bends downward, and we say that this function is concave downward.

• Unofficially, we can say a function is concave up on an interval $$I$$ if its graph over $$I$$ “holds water” and is concave down if the graph “spills water.”

 (a) Graph of an increasing function that bends upward (is concave up) (b) Graph of an increasing function that bends downward (is concave down)

In Figure 4.10, the graph of the function is concave upward. It is clear from this figure that in passing along the graph from left to right, the slope of the tangent line (which is $$f'(x)$$) increases (before $$P$$ the slope is negative, and after $$P$$ slope is positive; thus the tangent lines have increasing slopes).

In Figure 4.11, the graph of the function is concave downward; in passing along the graph from left to right, the slope of the tangent line (or $$f'(x)$$) decreases.

4.1. Assume $$f$$ is a differentiable function on an interval $$I$$.
(a) The (graph of the) function is “concave up” if $$f’$$ is increasing on $$I$$
(b) The (graph of the) function is “concave down” if $$f’$$ is decreasing on $$I$$.

• Because $$f”$$ is the derivative of $$f’$$, it follows from the Increasing/Decreasing Test (Theorem 4.1) that $$f’$$ is increasing on an interval $$I$$ if $$f”(x)>0$$ for each $$x$$ in $$I$$ and $$f’$$ is decreasing on $$I$$ if $$f”(x)<0$$ for each $$x$$ in $$I$$. Therefore, we have the following theorem.

4.4. Concavity Test. Let $$f$$ be a function whose second derivative exists at each point of an interval $$I$$.
(a) If $$f”(x)>0$$ for every $$x$$ in $$I$$, then the graph of $$f$$ is concave up on $$I$$.
(b) If $$f”(x)<0$$ for every $$x$$ in $$I$$, then the graph of $$f$$ is concave down on $$I$$.

Example 4.5. Determine where $$y=e^{-x^{2}}$$ is concave up and where it is concave up.

Solution

Let $$u=-x^{2}$$

$y=e^{-x^{2}}=e^{u}$ $\Rightarrow y’=e^{u}\frac{du}{dt}=e^{-x^{2}}(-2x)$ $y’=\underbrace{-2x}_{v}\underbrace{e^{-x^{2}}}_{w}$ \begin{aligned} \Rightarrow y” & =v’w+vw’\\ & =(-2)e^{-x^{2}}+(-2x)\left(\frac{d}{dx}e^{-x^{2}}\right)\\ & =-2e^{-x^{2}}+(-2x)\left(-2xe^{-x^{2}}\right)\\ & =-2e^{-x^{2}}+4x^{2}e^{-x^{2}}\\ & =4e^{-x^{2}}\left(x^{2}-\frac{1}{2}\right)\end{aligned} Because $$e^{-x^{2}}>0$$ for all $$x$$, the sign of $$y”$$ is solely determined by $$(x^{2}-1/2)$$. We write

$x^{2}-\frac{1}{2}=x^{2}-\left(\frac{1}{\sqrt{2}}\right)^{2}=\left(x-\frac{1}{\sqrt{2}}\right)\left(x+\frac{1}{\sqrt{2}}\right)$ [Recall $$A^{2}-B^{2}=(A-B)(A+B)$$.]

Therefore, we have the following sign table.

$$x$$ $$-\infty$$ $$-\frac{1}{\sqrt{2}}$$ $$\frac{1}{\sqrt{2}}$$ $$+\infty$$
sign of $$x-\frac{1}{\sqrt{2}}$$ $$-$$ $$-$$ $$-$$ $$0$$ $$+$$
sign of $$x+\frac{1}{\sqrt{2}}$$ $$-$$ $$0$$ $$+$$ $$-$$ $$+$$
$$\therefore$$ sign of $$y”$$ $$+$$ $$0$$ $$-$$ $$0$$ $$+$$
concavity Up Down Up

The graph of this function is concave down on the interval $$(-1/\sqrt{2},1/\sqrt{2})$$ and is concave up on $$(-\infty,-1/\sqrt{2})\cup(1/\sqrt{2},+\infty)$$. The graph of this function is shown in Figure 4.12.

### 4.4.2 Inflection points

Most curves are concave up on some interval and concave down on some others. A point where the direction of concavity changes is called an “inflection7 point” (see Figure 4.13)

Assume $$P(x_{0},f(x_{0}))$$ is an inflection point of the graph of $$f$$. If $$f”(x)$$ is continuous, because $$f”$$ has opposite signs at each side of $$P$$, then $$f”(x_{0})$$ must be zero. Even if the assumption of continuity of $$f”$$ is dropped, with a more advanced argument we can show that if $$f”(x_{0})$$ exists, it must be zero. The other possibility is $$f”(x_{0})$$ does not exist (for example, in Examples [exa:Ch6-Inflection-Ex2] and [exa:Ch6-Inflection-ex3]). So, we have the following theorem.

4.5. If $$(x_{0},f(x_{0}))$$ is an inflection point of the graph of $$f$$, either $$f”(x_{0})=0$$ or $$f”(x_{0})$$ does not exist.

For instance, from the last example, $$\left(\frac{1}{\sqrt{2}},e^{-1/2}\right)$$ and $$\left(-\frac{1}{\sqrt{2}},e^{-1/2}\right)$$ are the inflection points of $$y=e^{-x^{2}}$$.

• The above theorem states that if $$f$$ has an inflection point at $$x_{0}$$, then $$x_{0}$$ is a critical number of the derivative $$f’$$.

• Even if $$f”(x_{0})=0$$, the point $$(x_{0},f(x_{0}))$$ is not necessarily an inflection point. For instance, if $$f(x)=x^{4}$$, then $$f^{\prime\prime}(x)=12x^{2}$$ and $$f^{\prime\prime}(0)=0$$ but obviously $$f$$ has no inflection point.

1. Oxford Dictionary: from Latin inflectere, from in- ‘into’ + flectere ‘to bend’.↩︎