5 Steps to a 5 AP Calculus BC 2019

Limits and Continuity 65

Example 2

Determine the intervals on which the given function is continuous:

f(x)=

⎧

⎨

⎩

x^2 + 3 x− 10

x− 2

, x/= 2

10, x= 2

.

Check the three conditions of continuity atx=2:

Condition 1:f(2)=10.

Condition 2: limx→ 2

x^2 + 3 x− 10

x− 2

=xlim→ 2

(x+5)(x−2)

x− 2

=xlim→ 2 (x+5)=7.

Condition 3:f(2)/=limx→ 2 f(x). Thus, f(x) is discontinuous atx=2.

The function is continuous on (−∞, 2) and (2,∞). Verify your result with a calculator.

(See Figure 5.3-2.)

[–8,12] by [–3,17]

Figure 5.3-2

TIP • Remember that
d
dx

(

1

x

)

=−

1

x^2

and

∫

1

x

dx=ln|x|+C.

Example 3

For what value ofkis the function f(x)=

{

x^2 − 2 x, x≤ 6

2 x+k, x> 6

continuous atx=6?

Forf(x) to be continuous atx=6, it must satisfy the three conditions of continuity.

Condition 1:f(6)= 62 −2(6)=24.

Condition 2: limx→ 6 −(x^2 − 2 x)=24; thus limx→ 6 −(2x+k) must also be 24 in order for the limx→ 6 f(x)

to equal 24. Thus, limx→ 6 −(2x+k)=24, which implies 2(6)+k=24 andk=12. Therefore,

ifk=12,

Condition (3):f(6)=limx→ 6 f(x) is also satisfied.

Example 4

Givenf(x) as shown in Figure 5.3-3, (a) findf(3) and limx→ 3 f(x), and (b) determine iff(x)

is continuous atx=3. Explain your answer.

The graph off(x) shows thatf(3)=5 and the limx→ 3 f(x)=1. Sincef(3)=/xlim→ 3 f(x),f(x)

is discontinuous atx=3.