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

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

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

d[Differentiate] to find f′(x): f′(x)=x^2 −x−2. Set f′(x)= 0 ⇒x^2 −x− 2 =0or

(x−2)(x+1)=0.

Thus,x=2orx=−1, which implies f′(2)=0 andf′(−1)=0. Therefore, the values ofc

are−1 and 2. (See Figure 8.1-4.)

[–8, 8] by [–4, 4]

Figure 8.1-4

Example 3

The pointsP(1, 1) andQ(3, 27) are on the curve f(x)=x^3. Using the Mean Value

Theorem, findc in the interval (1, 3) such that f′(c) is equal to the slope of the

secantPQ.

The slope of secantPQism=

27 − 1

3 − 1

=13. Since f(x) is defined for all real numbers,

f(x) is continuous on [1, 3]. Alsof′(x)= 3 x^2 is defined for all real numbers. Thus, f(x)

is differentiable on (1, 3). Therefore, there exists a numbercin (1, 3) such that f′(c)=13.

Setf′(c)= 13 ⇒3(c)^2 =13 orc^2 =

13

3

,c=±

√

13

3

. Since only

√

13

3

is in the interval

(1, 3),c=

√

13

3

. (See Figure 8.1-5.)

[–4, 4] by [–20, 40]

Figure 8.1-5

Example 4

Let f be the function f(x)=(x−1)^2 /^3. Determine if the hypotheses of the Mean Value

Theorem are satisfied on the interval [0, 2], and if so, find all values ofcthat satisfy the

conclusion of the theorem.

Entery 1 =(x−1)^2 /^3. The graphy1 shows that there is a cusp atx=1. Thus, f(x)isnot

differentiable on (0, 2), which implies there may or may not exist acin (0, 2) such that

f′(c)=

f(2)−f(0)

2 − 0

. The derivative f′(x)=

2

3

(x−1)−^1 /^3 and

f(2)− f(0)

2 − 0

=

1 − 1

2

=0.