5 Steps to a 5 AP Calculus AB 2019 - William Ma

MA 3972-MA-Book April 11, 2018 17:21

160 STEP 4. Review the Knowledge You Need to Score High

(c) The functionf is increasing on (−∞,−4], [2, 4], and [8,∞) and is decreasing on

[−4, 2] and [4, 8].

(d) Summarize the information off′′on a number line:

concave

downward

concave

downward

– (^136)
++
concave
upward
concave
upward
decr. incr. decr. incr.
––
f′
f′′
f
A change of concavity occurs atx=−1, 3, and 6. Sincef′(x) exists, f has a tangent
at every point. Therefore,f has a point of inflection atx=−1, 3, and 6.
(e) The function f is concave upward on the interval (−1, 3) and (6,∞) and concave
downward on (−∞,−1) and (3, 6).
Example 4
A function f is continuous on the interval [−4, 3] with f(−4)=6 and f(3)=2 and the
following properties:

INTERVALS (−4,−2) X=−2(−2, 1) X= 1 (1, 3)

f′ − 0 − undefined +

f′′ + 0 − undefined −

(a) Find the intervals on which fis increasing or decreasing.

(b) Find where fhas its absolute extrema.

(c) Find wherefhas the points of inflection.

(d) Find the intervals where fis concave upward or downward.

(e) Sketch a possible graph off.

Solution:

(a) The graph of f is increasing on [1, 3] since f′>0 and decreasing on [−4,−2] and

[−2, 1] since f′<0.

(b) Atx=−4, f(x)=6. The function decreases untilx=1 and increases back to 2 at

x=3. Thus, f has its absolute maximum atx=−4 and its absolute minimum at

x=1.

(c) A change of concavity occurs atx=−2, and sincef′(−2)=0, which implies a tangent

line exists atx=−2,fhas a point of inflection atx=−2.

(d) The graph of f is concave upward on (−4, −2) and concave downward on

(−2, 1) and (1, 3).

(e) A possible sketch offis shown in Figure 8.4-5.