CHAPTER 11
11.30
The mean value theorem ensures the existence of a certain point on the graph of
(125, 5). Find the ^-coordinate of the point.
between (27,3) and
72
By the mean value theorem, there is a number c between 27 and 125 such that
11.31
Show that g(*) = Bx
3 - 6x
2 - 2x + 1 has a zero between 0 and 1.
Notice that the intermediate value theorem does not help, since g(0) = 1 and g(l) = l. Let f(x) =
2x
4 -2x
3 -x
2 + x and note that /'(*) = g(x). Since /(O) = /(I) = 0, Rolle's theorem applies to /(*) on
the interval [0,1]. Hence, there must exist c between 0 and 1 such that f'(c) = 0. Then g(c) = 0.
11.32
Show that x + 2x - 5 = 0 has exactly one real root.
Let f(x) = x
3 + 2x - 5. Since /(0)=-5<0 and /(2)=7>0, the intermediate value theorem tells us
that there is a root of f(x) = 0 between 0 and 2. Since f'(x) = 3x
2 + 2 > 0 for all x, f(x) is an increasing
function and, therefore, can assume the value 0 at most once. Hence, f(x) assumes the value 0 exactly once.
11.33
Suppose that f(x) is differentiable every where, that /(2) =-3, and that ! that 0 By the mean value theorem, there exists a c between 2 and 5 such that
So,
Since 2
Use the mean value theorem to prove that tan x > x for 0 < x < Tr/2.
The mean value theorem applies to tan* on the interval [0, x]. Hence, there exists c between 0 and x
such that sec
2 c = (tan* - tanO)/(*-0) = tan*/*. [Recall that D x (tan *) = sec
2 *.] Since 0
2 c > 1. Thus, tan xlx > 1, and, therefore, tan x> x.
11.35
If f'(x) = 0 throughout an interval [a, b], prove that /(*) is constant on that interval.
Let «<*
between a and * such that
throughout the interval.
11.36 If f'(x) = g'(x) for all x in an interval [a, b], show that there is a constant K such that f(x) = g(x) + K for
all x in [a, b].
Let h(x)=f(x)-g(x). Then h'(x) = 0 for all x in [a, £>]. By Problem 11.35, there is a constant K such
that h(x) = K for all x in [a, b]. Hence, /(x) = g(x) + K for all x in [a, 6].
11.37
Prove that x
3 + px + q = 0 has exactly one real root if p>0.
Let f(x) = x3 + px + q. Then /'(*) = 3*2 + p > 0. Hence, f(x) is an increasing function. So f(x) assumes the value 0 at most once. Now, lim f(x) = +°° and lim f(x) = —». Hence, there are numbers «
and i> where /(w) > 0 and f(v) < 0. By the intermediate value theorem, f(x) assumes the value 0 for some
number between u and v. Thus, f(x) has exactly one real root.
11.38
Prove the following generalized mean value theorem: If f(x) and g(x) are continuous on [a, b], and if fix) and
g(x) are differentiable on (a, b) with g'(x) ^ 0, then there exists a c in (a, b) such that
g(°) * g(b)- [Otherwise, if g(a) = g(b) = K, then Rolle's theorem applied to g(x) - K would yield a
number between a and b at which g'(x) = 0, contrary to our hypothesis.] Let
and set
F
(X) = f(x) ~ f(b> ~ L[g(x) ~ g(b)\. It is easy to see that Rolle's theorem applies to F(x). Therefore, there is a
number c between a and b for which F'(c) = 0. Then, /'(c) — Lg'(c) = 0, and
So,
/(5) + 3 = 3/'(c)
and 0
11.30
The mean value theorem ensures the existence of a certain point on the graph of
(125, 5). Find the ^-coordinate of the point.
between (27,3) and
72
By the mean value theorem, there is a number c between 27 and 125 such that
11.31
Show that g(*) = Bx
3 - 6x
2 - 2x + 1 has a zero between 0 and 1.
Notice that the intermediate value theorem does not help, since g(0) = 1 and g(l) = l. Let f(x) =
2x
4 -2x
3 -x
2 + x and note that /'(*) = g(x). Since /(O) = /(I) = 0, Rolle's theorem applies to /(*) on
the interval [0,1]. Hence, there must exist c between 0 and 1 such that f'(c) = 0. Then g(c) = 0.
11.32
Show that x + 2x - 5 = 0 has exactly one real root.
Let f(x) = x
3 + 2x - 5. Since /(0)=-5<0 and /(2)=7>0, the intermediate value theorem tells us
that there is a root of f(x) = 0 between 0 and 2. Since f'(x) = 3x
2 + 2 > 0 for all x, f(x) is an increasing
function and, therefore, can assume the value 0 at most once. Hence, f(x) assumes the value 0 exactly once.
11.33
Suppose that f(x) is differentiable every where, that /(2) =-3, and that ! that 0 By the mean value theorem, there exists a c between 2 and 5 such that
So,
Since 2
Use the mean value theorem to prove that tan x > x for 0 < x < Tr/2.
The mean value theorem applies to tan* on the interval [0, x]. Hence, there exists c between 0 and x
such that sec
2 c = (tan* - tanO)/(*-0) = tan*/*. [Recall that D x (tan *) = sec
2 *.] Since 0
2 c > 1. Thus, tan xlx > 1, and, therefore, tan x> x.
11.35
If f'(x) = 0 throughout an interval [a, b], prove that /(*) is constant on that interval.
Let «<*
between a and * such that
throughout the interval.
11.36 If f'(x) = g'(x) for all x in an interval [a, b], show that there is a constant K such that f(x) = g(x) + K for
all x in [a, b].
Let h(x)=f(x)-g(x). Then h'(x) = 0 for all x in [a, £>]. By Problem 11.35, there is a constant K such
that h(x) = K for all x in [a, b]. Hence, /(x) = g(x) + K for all x in [a, 6].
11.37
Prove that x
3 + px + q = 0 has exactly one real root if p>0.
Let f(x) = x3 + px + q. Then /'(*) = 3*2 + p > 0. Hence, f(x) is an increasing function. So f(x) assumes the value 0 at most once. Now, lim f(x) = +°° and lim f(x) = —». Hence, there are numbers «
and i> where /(w) > 0 and f(v) < 0. By the intermediate value theorem, f(x) assumes the value 0 for some
number between u and v. Thus, f(x) has exactly one real root.
11.38
Prove the following generalized mean value theorem: If f(x) and g(x) are continuous on [a, b], and if fix) and
g(x) are differentiable on (a, b) with g'(x) ^ 0, then there exists a c in (a, b) such that
g(°) * g(b)- [Otherwise, if g(a) = g(b) = K, then Rolle's theorem applied to g(x) - K would yield a
number between a and b at which g'(x) = 0, contrary to our hypothesis.] Let
and set
F
(X) = f(x) ~ f(b> ~ L[g(x) ~ g(b)\. It is easy to see that Rolle's theorem applies to F(x). Therefore, there is a
number c between a and b for which F'(c) = 0. Then, /'(c) — Lg'(c) = 0, and
So,
/(5) + 3 = 3/'(c)
and 0