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

MA 3972-MA-Book May 7, 2018 9:52

Limits and Continuity 101

[–3, 8] by [–4, 8]

Figure 6.3-3

Example 5

Ifg(x)=x^2 − 2 x−15, using the Intermediate Value Theorem show thatg(x) has a root in
the interval [1, 7].

Begin by findingg(1) andg(7), andg(1)=−16 andg(7)=20. Ifg(x) has a root, then
g(x) crosses thex-axis, i.e.,g(x)=0. Since− 16 ≤ 0 ≤20, by the Intermediate Value
Theorem, there exists at least one numbercin [1, 7] such thatg(c)=0. The numbercis a
root ofg(x).

Example 6

A functionfis continuous on [0, 5], and some of the values off are shown below.

x 0 3 5

f − 4 b − 4

Iff(x)=−2 has no solution on [0, 5], thenbcould be

(A) 1 (B) 0 (C) − 2 (D) − 5

Ifb=−2, thenx=3 would be a solution for f(x)=−2.

Ifb=0, 1, or 3, f(x)=−2 would have two solutions forf(x)=−2.

Thus,b=−5, choice (D). (See Figure 6.3-4.)
y

x

f(x) = – 2

3

2

1

• 10 12 345


• 2


• 3


• 5

(3, 1)

(3, 0)

(3, –2)

(0, –4) (3, –5) (5, –4)

Figure 6.3-4