### 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 $f''(3)=12>0$, 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 $f(x)=x^4-2x^2+11$.

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##### Calculus Refresher

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