# Concavity and Inflection Points

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### Inflection points, concavity upward and downward

An inflection points 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 $f”(3)=12>0$, 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

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