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) = x2 - 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, 3x2 = 1, x2 = 5, x = ±V5/3. The positive solution x = V5/3 lies between 0 and 1.
11.4 f(x) = 9x3-4x on [-§,§].
I f ' ( x ) = 27x2-4 and /(-§ )=/(§) = 0. Hence, Rolle's theorem is applicable. Setting f ' ( x ) = 0,
27x2 = 4, x2=£, * = ±2/3V3 = ±2V3/9. Both of these values lie in [-§, |], since 2V5/9<§.
11.5 /(*) = *3 - 3*2 + * + 1 on[l, 1 + V2].
I /'(*) = 3x2 - 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.
11.6 on [-2,3].
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]
11.7
if x = \
Notice that x3 -2x2 -5x + 6 = (x - l)(x2 - x -6). Hence, f(x) = x2 - x - 6 if x¥=l and x is in
[-2,3]. But /(*) = -6 = x2 - x - 6 when x = l. So f(x) = x2-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.
11.8 f(x) = x2/3~2x1':> on [0,8].
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.
11.9 if
if
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.
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, 70 CHAPTER 11
11.10 State the mean value theorem.
If fix) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there is a
number c in (a, b) such that
In Problems 11.11 to 11.16, determine whether the hypotheses of the mean value theorem hold for the function
f(x) on the given interval, and, if they do, find a value c satisfying the conclusion of the theorem.
11.11 f(x) = 2x + 3 on [1,4].
f ' ( x ) = 2. Hence, the mean value theorem applies. Note that Thus, we can
take c to be any point in (1,4).
11.12 f(x) = 3x2 - 5x + 1 on [2, 5].
/'(*) = 6*— 5, and the mean value theorem applies. Setting we
find which lies between 2 and 5.
11.13 f(x) = x3'4 on [0,16].
/(*) is continuous for x>0 and differentiable for x>0. Thus, the mean value theorem is applicable.
Setting we find which lies between 0 and 16.
11.14 on
Since is differentiable and nonzero on [1,3], f(x) is differentiable on [1,3].
Setting we obtain
The value lies between 1 and 3.
11.15 on
f(x) is differentiable on since on that interval. Setting
we obtain
Both of these values lie in [-3,4].
11.16 on
Since x-4 is differentiable and nonzero on [0, 2], so is/(*). Setting
we obtain The value
lies between 0 and 2.
11.17 Prove that, if f'(x)>0 for all x in the open interval (a, b), then f(x) is an increasing function on (a, b).
Assume a<u<v<b. Then the mean value theorem applies to f(x) on the closed interval (u, v). So, for
some c between u and v, f'(c) = [f(v) - /(«)] /(v - u). Hence, f(v) - /(«) = f'(c)(v - u). Since u<v,
v-u>0. By hypothesis, /'(c)>0. Hence, f(v) -/(«)>0, and /(u) >/(«). Thus,/(A:) is increasing
in (a, ft).
In Problems 11.18 to 11.26, determine where the function/is increasing and where it is decreasing.
11.18 f(x) = 3x + l.
f ' ( x ) = 3. Hence, f(x) is increasing everywhere.
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) = x2 - 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, 3x2 = 1, x2 = 5, x = ±V5/3. The positive solution x = V5/3 lies between 0 and 1.
11.4 f(x) = 9x3-4x on [-§,§].
I f ' ( x ) = 27x2-4 and /(-§ )=/(§) = 0. Hence, Rolle's theorem is applicable. Setting f ' ( x ) = 0,
27x2 = 4, x2=£, * = ±2/3V3 = ±2V3/9. Both of these values lie in [-§, |], since 2V5/9<§.
11.5 /(*) = *3 - 3*2 + * + 1 on[l, 1 + V2].
I /'(*) = 3x2 - 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.
11.6 on [-2,3].
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]
11.7
if x = \
Notice that x3 -2x2 -5x + 6 = (x - l)(x2 - x -6). Hence, f(x) = x2 - x - 6 if x¥=l and x is in
[-2,3]. But /(*) = -6 = x2 - x - 6 when x = l. So f(x) = x2-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.
11.8 f(x) = x2/3~2x1':> on [0,8].
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.
11.9 if
if
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
, 70 CHAPTER 11
11.10 State the mean value theorem.
If fix) is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there is a
number c in (a, b) such that
In Problems 11.11 to 11.16, determine whether the hypotheses of the mean value theorem hold for the function
f(x) on the given interval, and, if they do, find a value c satisfying the conclusion of the theorem.
11.11 f(x) = 2x + 3 on [1,4].
f ' ( x ) = 2. Hence, the mean value theorem applies. Note that Thus, we can
take c to be any point in (1,4).
11.12 f(x) = 3x2 - 5x + 1 on [2, 5].
/'(*) = 6*— 5, and the mean value theorem applies. Setting we
find which lies between 2 and 5.
11.13 f(x) = x3'4 on [0,16].
/(*) is continuous for x>0 and differentiable for x>0. Thus, the mean value theorem is applicable.
Setting we find which lies between 0 and 16.
11.14 on
Since is differentiable and nonzero on [1,3], f(x) is differentiable on [1,3].
Setting we obtain
The value lies between 1 and 3.
11.15 on
f(x) is differentiable on since on that interval. Setting
we obtain
Both of these values lie in [-3,4].
11.16 on
Since x-4 is differentiable and nonzero on [0, 2], so is/(*). Setting
we obtain The value
lies between 0 and 2.
11.17 Prove that, if f'(x)>0 for all x in the open interval (a, b), then f(x) is an increasing function on (a, b).
Assume a<u<v<b. Then the mean value theorem applies to f(x) on the closed interval (u, v). So, for
some c between u and v, f'(c) = [f(v) - /(«)] /(v - u). Hence, f(v) - /(«) = f'(c)(v - u). Since u<v,
v-u>0. By hypothesis, /'(c)>0. Hence, f(v) -/(«)>0, and /(u) >/(«). Thus,/(A:) is increasing
in (a, ft).
In Problems 11.18 to 11.26, determine where the function/is increasing and where it is decreasing.
11.18 f(x) = 3x + l.
f ' ( x ) = 3. Hence, f(x) is increasing everywhere.
