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

(Marvins-Underground-K-12) #1
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.

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