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Sullivan 04 apcalc4e 45342 ch02 166 233 5pp August 7, 2023 12:54

186 Chapter 2 • The Derivative and Its Properties

y Since lim f (x) = lim f (x),then lim f (x)exists.But lim f (x) = 8and f (3) = 5,so f
x→3 − x→3 + x→3 x→3
12
is discontinuous at 3. From the corollary, since f is discontinuous at 3, the function f
is not differentiable at 3.
8
Figure 20 shows the graph of f .
4 (3, 5)
NOW WORK Problem 43.
Determining Whether a Function Is DifferentiableNot Copy.
24 22 2 4 x
24
In Example 9, the function f is discontinuous at 3, so by the corollary, the derivative
of f at 3 does not exist. But when a function is continuous at a number c, then sometimes
2x + 2 if x < 3 the derivative at c exists and other times the derivative at c does not exist.
(
Figure 20 f (x) = 5 if x = 3
Figure 22 g(x) =BFW Publishers PAGES NOT FINAL - For Review Purposes Only - Do
2
x − 1 if x > 3
EXAMPLE 10
at a Number
Determine whether each piecewise-defined function is differentiable at c. If the function
has a derivative at c, find it.
3
x if x < 0 1 − 2x if x ≤ 1
(a) f (x) = 2 c = 0 (b) g(x) = c = 1
x if x ≥ 0 x − 2 if x > 1
Solution
y
4
(a) The function f is continuous at 0, which you should verify. To determine whether f
has a derivative at 0, we examine the one-sided limits at 0 using Form (1).
2
For x < 0,
(0, 0)
3
22 21 1 2 x f (x) − f (0) x − 0 2
lim = lim = lim x = 0
22 x→0 − x − 0 x→0 − x x→0 −
For x > 0,
24
2
f (x) − f (0) x − 0

x 3 if x < 0 lim = lim = lim x = 0
Figure 21 f (x) = 2 x→0 + x − 0 x→0 + x x→0 +
x if x ≥ 0
′
Since both one-sided limits are equal, f is differentiable at 0, and f (0) = 0. See
Figure 21.
y (b) The function g is continuous at 1, which you should verify. To determine whether g
6
is differentiable at 1, examine the one-sided limits at 1 using Form (1).
4 For x < 1,
g(x) − g(1) (1 − 2x) − (−1) 2 − 2x
2
lim = lim = lim
x→1 − x − 1 x→1 − x − 1 x→1 − x − 1
24 22 2 4 x = lim −2(x − 1) = lim (−2) = −2
(1, 21) x→1 − x − 1 x→1 −
22
For x > 1,
1 − 2x if x ≤ 1
© 2024 x − 2 if x > 1 x→1 + g(x) − g(1) = lim + (x − 2) − (−1) = lim + x − 1 = 1
lim
x − 1
x→1 x − 1
x − 1
x→1
The one-sided limits are not equal, so lim g(x) − g(1) does not exist. That is, g is not
x→1 x − 1
differentiable at 1. See Figure 22.
Notice in Figure 21 the tangent lines to the graph of f turn smoothly around the
origin. On the other hand, notice in Figure 22 the tangent lines to the graph of g change
abruptly at the point (1, −1), where the graph of g has a corner.
R
NOW WORK Problem 41 and AP Practice Problems 3, 4, 6, and 7.

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