Math Insight

Inflection points, concavity upward and downward

 

A point of inflection of the graph of a function $f$ is a point where the second derivative $f''$ is $0$. We have to wait a minute to clarify the geometric meaning of this.

A piece of the graph of $f$ is concave upward if the curve ‘bends’ upward. For example, the popular parabola $y=x^2$ is concave upward in its entirety.

A piece of the graph of $f$ is concave downward if the curve ‘bends’ downward. For example, a ‘flipped’ version $y=-x^2$ of the popular parabola is concave downward in its entirety.

The relation of points of inflection to intervals where the curve is concave up or down is exactly the same as the relation of critical points to intervals where the function is increasing or decreasing. That is, the points of inflection mark the boundaries of the two different sort of behavior. Further, only one sample value of $f''$ need be taken between each pair of consecutive inflection points in order to see whether the curve bends up or down along that interval.

Expressing this as a systematic procedure: to find the intervals along which $f$ is concave upward and concave downward:

  • Compute the second derivative $f''$ of $f$, and solve the equation $f''(x)=0$ for $x$ to find all the inflection points, which we list in order as $x_1 < x_2 <\ldots < x_n$. (Any points of discontinuity, etc., should be added to the list!)
  • We need some auxiliary points: To the left of the leftmost inflection point $x_1$ pick any convenient point $t_o$, between each pair of consecutive inflection points $x_i,x_{i+1}$ choose any convenient point $t_i$, and to the right of the rightmost inflection point $x_n$ choose a convenient point $t_n$.
  • Evaluate the second derivative $f''$ at all the auxiliary points $t_i$.
  • Conclusion: if $f''(t_{i+1})>0$, then $f$ is concave upward on $(x_i,x_{i+1})$, while if $f''(t_{i+1}) < 0$, then $f$ is concave downward on that interval.
  • Conclusion: on the ‘outside’ interval $(-\infty,x_o)$, the function $f$ is concave upward if $f''(t_o)>0$ and is concave downward if $f''(t_o) < 0$. Similarly, on $(x_n,\infty)$, the function $f$ is concave upward if $f''(t_n)>0$ and is concave downward if $f''(t_n) < 0$.

Example 1

Find the inflection points and intervals of concavity up and down of $$f(x)=3x^2-9x+6$$ First, the second derivative is just $f''(x)=6$.

Solution: Since this is never zero, there are not points of inflection. And the value of $f''$ is always $6$, so is always $>0$, so the curve is entirely concave upward.

Example 2

Find the inflection points and intervals of concavity up and down of $$f(x)=2x^3-12x^2+4x-27.$$

Solution: First, the second derivative is $f''(x)=12x-24$. Thus, solving $12x-24=0$, there is just the one inflection point, $2$. Choose auxiliary points $t_o=0$ to the left of the inflection point and $t_1=3$ to the right of the inflection point. Then $f''(0)=-24<0$, so on $(-\infty,2)$ the curve is concave downward. And $f''(2)=12>0$, so on $(2,\infty)$ the curve is concave upward.

Example 3

Find the inflection points and intervals of concavity up and down of $$f(x)=x^4-24x^2+11.$$

Solution: The second derivative is $f''(x)=12x^2-48$. Solving the equation $12x^2-48=0$, we find inflection points $\pm 2$. Choosing auxiliary points $-3,0,3$ placed between and to the left and right of the inflection points, we evaluate the second derivative: First, $f''(-3)=12\cdot 9-48>0$, so the curve is concave upward on $(-\infty,-2)$. Second, $f''(0)=-48 <0$, so the curve is concave downward on $(-2,2)$. Third, $f''(3)=12\cdot 9-48>0$, so the curve is concave upward on $(2,\infty)$.

Exercises

  1. Find the inflection points and intervals of concavity up and down of $f(x)=3x^2-9x+6$.
  2. Find the inflection points and intervals of concavity up and down of $f(x)=2x^3-12x^2+4x-27$.
  3. Find the inflection points and intervals of concavity up and down of $f(x)=x^4-2x^2+11$.