CHAPTER 11
Rolle's Theorem,
11.18 f(x) = 3x + l.
f ' ( x ) = 3. Hence, f(x) is increasing everywhere.
In Problems 11.18 to 11.26, determine where the function/is increasing and where it is decreasing.
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).
11.16
Since x-4 is differentiable and nonzero on [0, 2], so is/(*). Setting The value lies between 0 and 2.
Both of these values lie in [-3,4].
on that interval.
f(x) is differentiable on since Setting
we obtain 11.15
we obtain lies between 1 and 3.
Setting The value
is differentiable and nonzero on [1,3], f(x) is differentiable on [1,3].
Since
we obtain 11.14
/(*) is continuous for x>0 and differentiable for x>0. Thus, the mean value theorem is applicable.
we find Setting
11.13 f(x) = x3'4 on [0,16].
11.12 f(x) = 3x2 - 5x + 1 on [2, 5].
/'(*) = 6*— 5, and the mean value theorem applies. Setting which lies between 2 and 5.
find
we 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).
which lies between 0 and 16.
on
on
on
11.19 f(x) = -2x + 1.
I f ' ( x ) = —2 <0. Hence, f(x) is always decreasing.
11.20 f(x) = x2-4x + 7.
I f ' ( x ) = 2x-4. Since 2x-4>Q**x>2, f(x) is increasing when x>2. Similarly, since 2x-4<
0 <-» * < 2, f(x) is decreasing when x<2.
11.21 f(x) = 1 - 4x - x2.
1 f'(x)=-4-2x. Since -4- 2x>Q++x< -2, f(x) is increasing when x<-2. Similarly, f(x) is de-creasing when x > —2.
11.22 /(*) = Vl - x2.
f(x) is denned only for -1<*<1. Now, f'(x) = -xNl - x2. So, f(x) >Q**x <0. Thus, /(*) is increasing when -1< x < 0. Similarly, f(x) is decreasing when 0 < x < 1.
11.23
I f(x) is defined only when -3<x<3. f'(x)= \(-x)N 9 - x2. So, /'(*)> 0«-»*<0. Thus, f(x) is increasing when — 3<;t<0 and decreasing when 0 < A c < 3 .
11.24 f(x) = x3- 9x2 + 15x - 3.
I f ' ( x ) = 3x2-l8x + 15 = 3(x-5)(x- 1). The key points are x = l and * = 5. f'(x)>Q when jc>5, /'(AC)<O for Kx<5, and /'(AC)>O when x<l. Thus, /(*) is increasing when x<\ or x>5, and it is decreasing when 1 < x < 5.
11.25 f(x) = x + l/x.
I f(x) is denned for x^O. f ' ( x ) = \-(\lx2). Hence, /'(*)<O-H>!< l/x\ which is equivalent to x2 <l.
Hence, f(x) is decreasing when — 1<AC<0 or 0<x<l, and it is increasing when AC>! or A C < — 1 . 11.26 f(x) = x3- \2x + 20.
I f ' ( x ) = 3x2 — l2 = 3(x — 2)(x + 2). The key points are x = 2 and x = —2. For Ac>2, f'(x)>0; for -2<Ac<2, f'(x)<0; for jc<-2, f'(x)>Q. Hence, f(x) is increasing when x>2 or x<-2, and it is decreasing for -2 < x < 2.
11.27 Let f(x) be a differentiable function such that f'(x)^0 for all AC in the open interval (a, b). Prove that there is at most one zero of f(x) in (a, b).
I Assume that there exist two zeros u and v off(x) in (a, b) with u<v. Then Rolle's theorem applies to /(AC) in the closed interval [u, v]. Hence, there exists a number c in (u, v) such that f ' ( c ) = 0. Since a<c<b, this contradicts the assumption that f'(x)^0 for all x in (a, b).
11.28 Consider the polynomial f(x) = 5x3 - 2x2 + 3x-4. Prove that f(x) has a zero between 0 and 1 that is the only zero of/(AC).
I /(0)=-4<0, and /(1) = 2>0. Hence, by the intermediate value theorem, f(x) = 0 for some x between 0 and 1. /'(*) = 15AC2 - 4AC -I- 3. By the quadratic formula, we see that/'(*) has no real roots and is, therefore, always positive. Hence, f(x) is an increasing function and, thus, can take on the value 0 at most once.
11.29 Let/(AC) and g(x) be differentiable functions such that /(a)sg(a) and f'(x)> g'(x) for all x. Show that f(x) > g(x) for all x > a.
1 The function h(x) = f(x) - g(x) is differentiable, /*(«)>0, and h'(x)>0 for all x. By the latter condition, h(x) is increasing, and, therefore, since /i(«)>0, h(x)>0 for all AC > a. Thus, /M>g(Ac) for all AC > a.
ROLLE'STHEOREM, THE MEAN VALUE THEOREM, AND THE SIGN OFTHE DERIVATIVE 71
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(*) = Bx3 - 6x2 - 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) = 2x4 -2x3-x2 + 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) = x3 + 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 ) = 3x2 + 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 !</'(*)< 2 if 2<x<5. Show that 0</(5)<3.
By the mean value theorem, there exists a c between 2 and 5 such that So, Since 2 < c < 5 , K/'(c)<2, 3<3/'(c)<6, 3</(5) + 3<6,
11.34 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 sec2 c = (tan* - tanO)/(*-0) = tan*/*. [Recall that Dx(tan *) = sec2 *.] Since 0 < c < 7 r / 2 , 0 < cos c < 1, sec c > 1, sec2 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 « < * < f c . The mean value theorem applies to/(*) on the interval [«,*]. Hence, there exists a c Since f ' ( c ) = 0, /(*)=/(«). Hence,/(*) has the value/(a) 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 x3 + 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) as-sumes 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</(5)<3
ROLLE'S THEOREM, THE MEAN VALUE THEOREM, AND THE SIGN OF THE DERIVATIVE
11.39 Use the generalized mean value theorem to show that
f Let f(x) = sinx and g(x) = x. Since e'(x) = l, the generalized mean value theorem applies to the
11.41 Apply the mean value theorem to the following functions on the interval [-1,8]. (a) f(x) = x4'3 (b) g(x) = x2'\
However, there is no number c in (-1,1) for which f ' ( c ) = 0, since /'(*) = 1 for x>0 and /'(.v)=-l for x<0. Of course,/'(O) does not exist, which
is the reason that the mean value theorem does not apply.
11.44 Find a point on the graph of y = x2+x + 3, between .v = 1 and x = 2, where the tangent line is parallel to the line connecting (1,5) and (2,9).
Hence, we must find c such that 2Ac + B = A(b + a) + B. Then c = \(b + a). Thus, the point is the midpoint of the interval.
73
Hence, there is a number c such that 0 < c < x for which interval [0, x] when x>0.
11.40 Show that |sin u — sin v\ s |« — u|.
By the mean value theorem, there exists a c between u and v for which Since
By the mean value theorem, there is a number c between -1 and 8 such that (b) The mean value theorem is not applic-able because g'(*) does not exist at x = 0.
11.42 Show that the equation 3 tan x + x* = 2 has exactly one solution in the interval [0, ir/4].
Let f(x) = 3 tan x + x3. Then /'(*) = 3 sec" x + 3x~ > 0, and, therefore, f(x) is an increasing function.
Thus, f(x) assumes the value 2 at most once. But, /(0) = 0 and /(Tr/4) = 3 + (ir/4)3 >2. So, by the intermediate value theorem, f(c) = 2 for some c between 0 and rr/4. Hence, f(x) = 2 for exactly one x in
[0, 7T/4].
11.43 Give an example of a function that is continuous on [ —1. 1] and for which the conclusion of the mean value theorem does not hold.
Then
This is essentially an application of the mean value theorem to f(x) = x~ + x + 3 on the interval [1, 21.
The slope of the line connecting (1,5) and (2,9) is For that line to be parallel to the tangent line at a point (c, /(c)), the slope of the tangent line, f ' ( c ) , must be equal to 4. But, /'(•*) =
2x + 1. Hence, we must have 2c + l = 4, c = § . Hence, the point is (|, ").
11.45 For a function f(x) = Ax2 + Bx + C, with A 7^0, on an interval [a, b], find the number in (a, b) determined by the mean value theorem.
f ' ( x ) = 2Ax + B. On the other hand,
11.46 If/is a differentiable function such that lim /'(.v) = 0, prove that lim [f(x + 1) -/(*)] = 0.
By the mean value theorem, there exists a c with x<c<x + l such that f(x+1) -/(*) = f'(c). As *-»+=», c-»+°°. Hence,/'(c) approaches 0, since lim f ' ( x ) = 0. Therefore, lim [f(x +
l)-/(jc)] = 0.
Let f(X) = \x\.
As we also have
and Hence,
Hence, Since
Hence,
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CHAPTER 11 11.47
11.48 An important function in calculus (the exponential function) may be defined by the conditions
Prove that the zeros of sin x and cos x separate each other; that is, between any two zeros of sin x, there is a zero of cos x, and vice versa.
Assume sina = 0 and sin 6 = 0 with a<b. By Rolle's theorem, there exists a c with a<c<b such that cosc = 0, since DA.(sin x) = cos x. Similarly, if coso = cosfe=0 with a<b, then there ex-ists a c with a<c<b such that sinc = 0, since D,,(cosjr) = — sin x.
Fig. 11-1 74
/'(*) = /« (~°° <*<+*) and /(0)=1 (1) Show that this function is (a) strictly positive, and (b) strictly increasing.
(a) Let h(x) =f(x)f(-x); then, using the product rule, the chain rule, and (/), h'(x)=f'(x)f(-x) + f(x)f'(-x)(-l)=f(x)f(-x)-f(x)f(-x) = 0. So by Problem 11.35, h(x) = const. = h(0) = 1-1 = 1; that is,
for all x,
/«/(-*) = !
By (2), f(x) is never zero. Furthermore, the continuous (because it is differentiable) function/(x) can never be negative; for f(a) < 0 and /(O) = 1 > 0 would imply an intermediate zero value, which we have just seen to be impossible. Hence f(x) is strictly positive. (b) /'(*) =/(*) >0; so (Problem 11.17), f(x) is strictly increasing.
11.49 Give an example of a continuous function f(x) on fO, 11 for which the conclusion of Rolle's theorem fails.
Let f(x)={-\x-{\ (see Fig. 11-1). Then /(O) =/(!) = 0, but f ' ( x ) is not 0 for any x in (0,1).
/'(*) = ±1 for all A: in (0,1), except at jc = { , where f ' ( x ) is not defined.
(2)