5 Steps to a 5 AP Calculus BC 2019

Graphs of Functions and Derivatives 105

Example 1

If f(x)=x^2 + 4 x−5, show that the hypotheses of Rolle’s Theorem are satisfied on the
interval [−4, 0] and find all values ofcthat satisfy the conclusion of the theorem. Check the
three conditions in the hypothesis of Rolle’s Theorem:

(1) f(x)=x^2 + 4 x−5 is continuous everywhere since it is polynomial.
(2) The derivativef′(x)= 2 x+4 is defined for all numbers and thus is differentiable on
(−4, 0).
(3) f(0)= f(−4)=−5. Therefore, there exists acin (−4, 0) such that f′(c)=0. To find
c, set f′(x)=0. Thus, 2x+ 4 = 0 ⇒x=−2, i.e.,f′(−2)=0. (See Figure 7.1-3.)

[–5,3] by [–15,10]

Figure 7.1-3

Example 2

Letf(x)=
x^3
3

−

x^2

2

− 2 x+2. Using Rolle’s Theorem, show that there exists a numbercin

the domain offsuch that f′(c)=0. Find all values ofc.

Note f(x) is a polynomial and thus f(x) is continuous and differentiable everywhere.

Entery 1 =
x^3
3

−

x^2

2

− 2 x+2. The zeros ofy1 are approximately− 2 .3, 0.9, and 2.9

i.e., f(− 2 .3)= f(0.9)= f(2.9)=0. Therefore, there exists at least onecin the interval
(− 2 .3, 0.9) and at least one c in the interval (0.9, 2.9) such that f′(c)=0. Use
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 7.1-4.)

[–8,8] by [–4,4]

Figure 7.1-4