5128_Ch04_pp186-260

Section 4.2 Mean Value Theorem 197
y
B(2, 4)
y = x 2
(1, 1)
A(0, 0) 1 2
Figure 4.12 (Example 1)
x
SOLUTION
The function f x x 2 is continuous on 0, 2 and differentiable on 0, 2. Since
f 0 0 and f 2 4, the Mean Value Theorem guarantees a point c in the interval
0, 2 for which
f c f b f a
b a
2c f 2 f 0
2
2 0
c 1.
fx 2x
Interpret The tangent line to f x x 2 at x 1 has slope 2 and is parallel to the
chord joining A0, 0 and B2, 4 (Figure 4.12).
EXAMPLE 2
Exploring the Mean Value Theorem
Now try Exercise 1.
Explain why each of the following functions fails to satisfy the conditions of the Mean
Value Theorem on the interval [–1, 1].
(a) f (x) x 2 1
(b)
x 3 3 for x 1
f x { x
2
1 for x 1
SOLUTION
(a) Note that x 2 1 |x| 1, so this is just a vertical shift of the absolute value
function, which has a nondifferentiable “corner” at x 0. (See Section 3.2.) The
function f is not differentiable on (–1, 1).
(b) Since lim x→1
– f (x) lim x→1
– x 3 3 4 and lim x→1
+ f (x) lim x→1
+ x 2 1 2, the
function has a discontinuity at x 1. The function f is not continuous on [–1, 1].
If the two functions given had satisfied the necessary conditions, the conclusion of the
Mean Value Theorem would have guaranteed the existence of a number c in (– 1, 1)
such that f(c) f (1 ) f (1)
0. Such a number c does not exist for the function in
1 ( 1)
part (a), but one happens to exist for the function in part (b) (Figure 4.13).
y
y
3
2
4
1
–2 –1 0
1 2
x
– 4
0
4
x
(a)
Figure 4.13 For both functions in Example 2, f (1 ) f (1)
0 but neither
1 ( 1)
function satisfies the conditions of the Mean Value Theorem on the interval
[– 1, 1]. For the function in Example 2(a), there is no number c such that
f(c) 0. It happens that f(0) 0 in Example 2(b).
(b)
Now try Exercise 3.