5 Steps to a 5 AP Calculus BC 2019

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

(b) Andfhas the smallest value atx=x 4.

(c)

A change of concavity occurs atx=x 3 ,

andf′(x 3 ) exists, which implies there

is a tangent tof atx=x 3. Thus, at

x=x 3 ,f has a point of inflection.

(d) The functionf′′represents the slope

of the tangent tof′. The slope of the

tangent tof′is the largest atx=x 4.

14. (a)Sincef′(x) represents the slope of the
    tangent, f′(x)=0atx=0, andx=5.
    (b)Atx=2,fhas a point of inflection,
    which implies that if f′′(x) exists,
    f′′(x)=0. Sincef′(x) is differentiable
    for all numbers in the domain,f′′(x)
    exists, andf′′(x)=0atx=2.
    (c)Since the function fis concave
    downward on (2,∞),f′′<0on
    (2,∞), which impliesf′is decreasing
    on (2,∞).


15. (a)The functionf is increasing on the
    intervals (−2, 1) and (3, 5) and
    decreasing on (1, 3).
    (b)The absolute maximum occurs at
    x=1, since it is a relative maximum,
    f(1)>f(−2) andf(5)<f(−2).
    Similarly, the absolute minimum
    occurs atx=3, since it is a relative
    minimum, andf(3)< f(5)<f(−2).

(c) No point of inflection. (Note that at

x=3,f has a cusp.)

Note: Some textbooks define a point

of inflection as a point where the

concavity changes and do not require

the existence of a tangent. In that case,

atx=3,f has a point of inflection.

(d) Concave upward on (3, 5) and

concave downward on (−2, 3).

(e) A possible graph is shown in

Figure 7.9-6.

Figure 7.9-6

16. (a)

decr incr decr

rel. min. rel. max.

• +–

f^06

f ′

The functionfhas its relative

minimum atx=0 and its relative

maximum atx=6.

(b) The functionfis increasing on [0, 6]

and decreasing on (−∞, 0] and [6,∞).

(c)

incr decr

+–

concave

upward

concave

downward

pt. of

inflection

3

f ′

f ′′

f