5 Steps to a 5 AP Calculus BC 2019

Graphs of Functions and Derivatives 117

Example 2

Using a calculator, find the values ofxat which the graph ofy=x^2 exchanges concavity.

Entery 1 =x∧ 2 ∗e∧xandy 2 =d(y1(x),x, 2). The graph ofy2, the second derivative
ofy, is shown in Figure 7.2-21. Using the [Zero] function, you obtainx=− 3 .41421 and
x=− 0 .585786. (See Figures 7.2-21 and 7.2-22.)

[–4,1] by [–2,5]

Figure 7.2-21

++–

–3.41421 –0.585786

x

f

f ′′

Concave

upward

Concave

downward

Concave

upward

Change of

concavity

Change of

concavity

Figure 7.2-22

Thus,fchanges concavity atx=− 3 .41421 andx=− 0 .585786.

Example 3

Find the points of inflection off(x)=x^3 − 6 x^2 + 12 x−8 and determine the intervals where
the functionfis concave upward and where it is concave downward.

Step 1: Find f′(x) andf′′(x).
f′(x)= 3 x^2 − 12 x+ 12
f′′(x)= 6 x− 12

Step 2: Setf′′(x)=0.
6 x− 12 = 0
x= 2
Note that f′′(x) is defined for all real numbers.

Step 3: Determine intervals.

The intervals are (−∞, 2) and (2,∞).