CHAPTER 11
Rolle's Theorem,
the Mean Value Theorem,
and the Sign of the Derivative
11.1
State Rolle's theorem.
f If / is continuous over a closed interval [a, b] and differentiable on the open interval (a, b), and if
/(a) = f(b) = 0, then there is at least one number c in (a, b) such that f'(c) = 0.
In Problems 11.2 to 11.9, determine whether the hypotheses of Rolle's theorem hold for the function/on the
given interval, and, if they do, verify the conclusion of the theorem.
11.2
f(x) = x
2 - 2x - 3 on [-1,3].
I f(x) is clearly differentiable everywhere, and /(-I) =/(3) = 0. Hence, Rolle's theorem applies. /'(*) =
2x-2. Setting /'(*) = °> we obtain x = l. Thus, /'(1) = 0 and -KK3.
11.3
/(*) = x" - x on [0,1].
I f(x) is differentiable, with /'(*) = 3*
2 -1. Also, /(0)=/(1) = 0. Thus, Rolle's theorem applies.
Setting /'(*) = 0, 3x
2 = 1, x
2 = 5, x = ±V5/3. The positive solution x = V5/3 lies between 0 and 1.
11.4
f(x) = 9x
3 -4x on [-§,§].
I f'(x) = 27x
2 -4 and /(-§ )=/(§) = 0. Hence, Rolle's theorem is applicable. Setting f'(x) = 0,
27x
2 = 4, x
2 =£, * = ±2/3V3 = ±2V3/9. Both of these values lie in [-§, |], since 2V5/9<§.
11.5
/(*) = *
3 - 3*
2 + * + 1 on[l, 1 + V2].
I /'(*) = 3x
2 - 6^ + 1 and /(I) =/(! + V2) = 0. This means that Rolle's theorem applies. Setting
f'(x) = 0 and using the quadratic formula, we obtain x = l±^V6 and observe that 1< 1 + jV~6< 1 + V2.
on [-2,3].
11.6
There is a discontinuity at * = !, since lim f(x) does not exist. Hence, Rolle's theorem does not apply.
X—»1
if x ¥= 1 and x is in [—2, 3]
if x = \
11.7
11.8
f(x) = x
2/3 ~2x
1 '
:> on [0,8].
11.9
f(x) is not differentiable at * = 1. (To see this, note that, when Ax<0, [/(! + Ax) - 1]/A* = 2 +
Ax-*2 as Ax-»0. But, when A*>0, [/(I + A*) - 1]/A* = -l-» -1 as Ax-»0.) Thus, Rolle's
theorem does not apply.
69
if
if
f(x) is differentiable within (0,8), but not at 0. However, it is continuous at x = 0 and, therefore,
throughout [0,8]. Also, /(0)=/(8) = 0. Hence, Rolle's theorem applies. /'(*) = 2/3v^-2/3(vT)
2 .
Setting f'(x) = 0, we obtain x = 1, which is between 0 and 8.
Notice that x
3 -2x
2 -5x + 6 = (x - l)(x
2 - x -6). Hence, f(x) = x
2 - x - 6 if x¥=l and x is in
[-2,3]. But /(*) = -6 = x
2 - x - 6 when x = l. So f(x) = x
2 -x-6 throughout the interval [-2, 3].
Also, note that /(-2) =/(3) = 0. Hence, Rolle's theorem applies. f'(x) = 2x-l. Setting f'(x) = 0, we
obtain x = 5 which lies between —2 and 3.
Rolle's Theorem,
the Mean Value Theorem,
and the Sign of the Derivative
11.1
State Rolle's theorem.
f If / is continuous over a closed interval [a, b] and differentiable on the open interval (a, b), and if
/(a) = f(b) = 0, then there is at least one number c in (a, b) such that f'(c) = 0.
In Problems 11.2 to 11.9, determine whether the hypotheses of Rolle's theorem hold for the function/on the
given interval, and, if they do, verify the conclusion of the theorem.
11.2
f(x) = x
2 - 2x - 3 on [-1,3].
I f(x) is clearly differentiable everywhere, and /(-I) =/(3) = 0. Hence, Rolle's theorem applies. /'(*) =
2x-2. Setting /'(*) = °> we obtain x = l. Thus, /'(1) = 0 and -KK3.
11.3
/(*) = x" - x on [0,1].
I f(x) is differentiable, with /'(*) = 3*
2 -1. Also, /(0)=/(1) = 0. Thus, Rolle's theorem applies.
Setting /'(*) = 0, 3x
2 = 1, x
2 = 5, x = ±V5/3. The positive solution x = V5/3 lies between 0 and 1.
11.4
f(x) = 9x
3 -4x on [-§,§].
I f'(x) = 27x
2 -4 and /(-§ )=/(§) = 0. Hence, Rolle's theorem is applicable. Setting f'(x) = 0,
27x
2 = 4, x
2 =£, * = ±2/3V3 = ±2V3/9. Both of these values lie in [-§, |], since 2V5/9<§.
11.5
/(*) = *
3 - 3*
2 + * + 1 on[l, 1 + V2].
I /'(*) = 3x
2 - 6^ + 1 and /(I) =/(! + V2) = 0. This means that Rolle's theorem applies. Setting
f'(x) = 0 and using the quadratic formula, we obtain x = l±^V6 and observe that 1< 1 + jV~6< 1 + V2.
on [-2,3].
11.6
There is a discontinuity at * = !, since lim f(x) does not exist. Hence, Rolle's theorem does not apply.
X—»1
if x ¥= 1 and x is in [—2, 3]
if x = \
11.7
11.8
f(x) = x
2/3 ~2x
1 '
:> on [0,8].
11.9
f(x) is not differentiable at * = 1. (To see this, note that, when Ax<0, [/(! + Ax) - 1]/A* = 2 +
Ax-*2 as Ax-»0. But, when A*>0, [/(I + A*) - 1]/A* = -l-» -1 as Ax-»0.) Thus, Rolle's
theorem does not apply.
69
if
if
f(x) is differentiable within (0,8), but not at 0. However, it is continuous at x = 0 and, therefore,
throughout [0,8]. Also, /(0)=/(8) = 0. Hence, Rolle's theorem applies. /'(*) = 2/3v^-2/3(vT)
2 .
Setting f'(x) = 0, we obtain x = 1, which is between 0 and 8.
Notice that x
3 -2x
2 -5x + 6 = (x - l)(x
2 - x -6). Hence, f(x) = x
2 - x - 6 if x¥=l and x is in
[-2,3]. But /(*) = -6 = x
2 - x - 6 when x = l. So f(x) = x
2 -x-6 throughout the interval [-2, 3].
Also, note that /(-2) =/(3) = 0. Hence, Rolle's theorem applies. f'(x) = 2x-l. Setting f'(x) = 0, we
obtain x = 5 which lies between —2 and 3.