f(xi)
=
f’(c)(x2
x)
—
>
I
). This is
1
f(x
=
F(x)
=
F’(x)
—
G’(x)
G(x) + C
G(x). Then
H’(x)
—
=
0
=
—
=
—
)
5
11(x
=
—
H’(c)(x
—
)
1
x
4. Since D(x
) = 4x
4
, it follows that every function F that
3
satisfies F’(x) = 4x
3 has the form F(x) =
3. If two functions F and G have the same derivative on the
interval (a, b), then there is a constant C such that
)
1
11(x
= 0 by hypothesis. Therefore, 11(x)
0 or, equivalently,
11(xi) for all x in (a, b). Since ff(x) = F(x) — G(x), we conclude that
G(x) = ff(x
). Now let C = H(xi), and we have the conclusion
1
I
G(x) + C.
then there is a point c
11(x)
F(x)
F(x)
But H’(c)
ff(x)
, as some (fixed) point in (a, b), and let x be any other
1
for all x in (a, b). Choose x
point there. The function H satisfies the hypotheses of the Mean Value Theorem on
, and x. Thus, there is a number c between x
1
1
the closed interval with end points x
and x such that
Proof Let 11(x)
F(x)
G’(x) for all x in (a, b), then there is a constant C such that
for all x in (a, b).
If F’(.x)
Theorem B
Our next theorem will be used repeatedly in this and the next chapter. In
words, it says that two functions with the same derivative differ by a constant, possi
bly the zero constant (see Figure 7).
2. The function, f(x) = I sin x I would satisfy the hypotheses
of the Mean Value Theorem on the interval [0, 1] but would fbi
satisfy
on [a, b] and differentiable on
in (a, b) such that
-
)
2
) > 0; that is, f(x
1
f(x
Since f’(c) > 0, we see that f(x
)
2
what we mean when we say thatf is increasing on I.
The case wheref’(x) < 0 on I is handled similarly.
f(’2)
Proof of the Monotoulcity Theorem We suppose that f is continuous on 1
and that f’(x) > 0 at each point x in the interior of I. Consider any two points x
1
. By the Mean Value Theorem applied to the interval
2
and x
2 of I with x < x
,x
1
) satisfying
7
,x
1
[x
], there is a number c in (x
2
In Section 3.2, we promised a rigorous proof of the
Monotonicity Theorem (Theorem 3.2A). This is the theorem that relates the sign
of the derivative of a function to whether that function is increasing or decreasing.
The Theorem Used
—
1. The Mean Value Theorem for Derivatives says that if f is
Concepts Review
As with most topics in this book, you
should try to see things from an
algebraic and a geometrical point of
view Geometrically, Theorem B says
that if F and C have the same deriv
ative then the graph of C is a verti
cal translation of the graph of F.
Geometry and Algebra
Figure 7
V
F
—
The average velocity over the interval [3, 6] is equal to
3) = 8. The instantaneous velocity is s’(t) = 2t — 1. To find
(s(6) — s(3))/(6
the point where average velocity equals instantaneous velocity, we equate
8 2t
1 andsoLve to gett = 9/2.
I
SOLUTION
188 Chapter 3 Applications of the Derivative
=
=
16. C(O)
18. f(x)
=
=
14. g(x)
20. f(x)
=
=
=
=
=
=
12. h(t)
10. f(x)
8. F(t)
7. f(z)
6. F(x)
5. H(s)
3. f(x)
+
z
—
[-2,2]
x]; [1,2]
21. f(x)
19. f(x)
+
!; [—i,]
x
17. T(O)
csc 9;
[—,
15. S(6)
ir]
; [—1, 1]
3
x’
11. h(t)
13. g(x)
[0, 4]
9. h(x)
4): [-1,2]
213 [—2,2]
t
;
;
—--j;[02]
(z
2
s
4. g(x)
1;[—3,1]
—
+
3s
x; [—2, 2]
+
2
x
=
=
=
+
x
x
+
+
xi; [-2,1]
tan 0; [0, ir]
sin 0; [—v-, ir]
[0, 1]
[0, 2]
5!3;
273
t
;
[0,2]
1); [—1, 11
—--;
(x
26. Use Problem 25 to show that each of the following is in
creasing on (—cc, cc).
(a) f(x) =
x
(b) f(x)
x0
3
Ix
(c) f(x)=
11
x>o
25. Prove: 1ff is continuous on (a, b) and if f’(x) exists and
satisfies f’(x) > 0 except at one point x
0 in (a, b), then f is in
creasing on (a, b). Hint: Consider f on each of the intervals (a, xol
and [x
, b) separately.
0
24. Show that if f is the quadratic function defined by
f(x)
2 + /3x + y, a
ax
0, then the number c of the Mean
Value Theorem is always the midpoint of the given interval [a, b].
Figure 8
y
orem for the interval [0,8].
23. For the function graphed in Figure 8, find (approximate
ly) all points c that satisfy the conclusion to the Mean Value The
22. (RolIe’s Theorem) 1ff is continuous on [a, b] and differen
tiable on (a, b), and if f(a) = f(b), then there is at least one nuin
ber c in (a, b) such that f’(c) = 0. Show that Rolle’s Theorem is
just a special case of the Mean Value Theorem. (Michel Rolle
(1652—1719) was a French mathematician.)
in
each of the Problems 1—21, a function is defined and a closed in
terval is given. Decide whether the Mean Value Theorem applies to
the given function on the given interval. If it does, find alt possible
values of c; if not, state the reason. In each problem, sketch the
graph of the given function on the given interval.
2. g(x) = xi; [—2,2]
1. f(x) = xi;[l,2]
problem Set 3.6
=
=
2 de
l/t
1/t de
189
=
4. Find a formula
—
If(x2)
-
f(xi)I
2
Mx
-
xii
38. Prove that if If’(x)i
M for all x in (a, b) and if x
1 and
are any two points in (a, b) then
—
41. Prove that, if f is continuous on land if f’(x) exists and
satisfies f’ (x)
0 on the interior of I, then f is nondecreasing on
I. Similarly, iff’(x)
0, then f is nonincreasing on I.
40. A functionf is said to be nondecreasing on an interval lit
x x
<
1
inl.Similarly,fisnonin2
f(xi)
2
)forx andx
2
f(x
1
1 and x
2 in I.
1 < x
creasingonlifx
f(x)
2
) forx
2
f(x
(a) Sketch the graph of a function that is nondecreasing but not
increasing.
(b) Sketch the graph of a function that is nonincreasing but not
decreasing.
39. Show that f(x) = sin 2x satisfies a Lipschitz condition
with constant 2 on the interval (—cc, cc). See Problem 38.
Note: A function satisfying the above inequality is said to satisfy
a Lipschitz condition with constant M. (Rudolph Lipschitz
(1832—1903) was a German mathematician.)
12
—
37. Let f(x) = (x
1)(x 2)(x 3). Prove by using
Problem 36 that there is at least one value in the interval [0, 4]
where f”(x) = 0 and two values in the same interval where
f’(x) = 0.
36. Let g be continuous on [a, b] and suppose that g”(x)
exists for all x in (a, b). Prove that if there are three values of x in
[a, b] for which g(x) = 0 then there is at least one value of x in
(a, b) such that g”(x) = 0.
—
34. Show that f(x) = 2x
3
2 + 1 = 0 has exactly one
9x
solution on each of the intervals (—1,0), (0, 1), and (4. 5). Hint:
Apply Problem 33.
35. Let f have a derivative on an interval I. Prove that
between successive distinct zeros of f’ there can be at most one
zero of f. Hint: Try a proof by contradiction and use Rolle’s
Theorem (Problem 22).
33. Prove: Let f be continuous on [a, b] and differentiable on
(a, b). If f(a) and f(b) have opposite signs and if f’ (x)
0 for all
x in (a, b), then the equation f(x) = 0 has one and only one solu
tion between a and b. Hint: Use the Intermediate Value Theorem
and Rolle’s Theorem (Problem 22).
32. Suppose that F’(x) = 5 and F(0)
for F(x). Hint: See Problem 31.
31. Prove that if F’(x) = D for all x in (a, b) then there is a
constant C such that F(x) = Dx + C for all x in (a, b). Hint: Let
G(x) = Dx and apply Theorem B.
30. Suppose that you know that cos(0) = 1, sin(0) = 0,
D cos x = —sin x. and D sin x = cos x, but nothing else about
the sine and cosine functions. Show that cos
2 x + sin
2 x = 1.
Hint: Let F(x) = cos
2 x + sin
2 x and use Problem 29.
29. Prove that if F’(x) = 0 for all x in (a, b) then there is a
constant C such that F(x) = C for all x in (a, b). Hint. Let
G(x) = OandapplyTheoremB.
28. Use the Mean Value Theorem to show that s
creases on any interval to the right of the origin.
27. Use the Mean Value Theorem to show that s
creases on any interval over which it is defined.
Section 3.6 The Mean Value Theorem for Derivatives
.r
g(x)
0 on L then
—
)
1
g(x
—
h(xj)
h’(x) for allx in (a, b) then
0 and f’(x)
2 is
f
.11
-
lim(\’x-i-2
—
=
0
—
sinyl
-
—
—
f(x)
M(
3.7
Solving Equations
Numerically
then f is a constant function.
49. Prove that if f(y)
.1)2
for all .r andy
-.4
1:
Y=X3r3;
x
Bisection Method
—
—
—
=
=
in,,
—
3
x
—
3x
.
0
b
in,,.
—
=
=
—
5
=
0 to accu
—
)/2
1
a
f(2.5)
2 and b
2
1
ni
2.5.
5
3.125
(—3)(3.125)
0.5
—
=
—9.375 <0
Next we increment n so that it has the value 2 and repeat these steps. We can
continue this process to obtain the entries in the following table:
=
2)/2
f(2)f(2.5)
—
32.5
2.5
Step 5: The condition f(a,,) f(m
) > 0 is false.
0
1
a
=
(3
—
(2 + 3)/2
2.5
=
sively bisecting an interval known to contain a solution. This Bisection Method has
two great virtues—simplicity and reliability. It also has a major vice—the large
number of steps needed to achieve the desired accuracy (otherwise known as slow
ness of convergence).
=
1
(b
=
)/2
1
1 + b
(a
)
1
f(a
)
1
f (in
Step 4: Since
=
)
1
f(m
Step 3: h
1
Step 2:
Step 1: m
1
SOLUTION We first sketch the graph of y
3
x
3x
5 (Figure 3) and,
noting that it crosses the x-axis between 2 and 3, we begin with a
1
2 and 1
3.
b
=
and b,,
1
a,, and b,,+
1
we set a
2
sJj
a,,)/2.
0, then set a,,+
—
• EXAMPLE 1 Determine the real root of f(x)
racy within 0.0000001.
.
<
(b
The Bisection Method In Example 7 of Section 1.6 we saw how to use the
Intermediate Value Theorem to approximate a solution of f(x)
0 by succes
Figure 3
/Th
If f(a,,) f(m) > 0, then set a,,
1
5.
.
=
If f(a,,) f(m,,)
0, then STOP
4.
=
Calculate h,,
and if f(m,,)
3.
f(rn,,),
(a,, + b,,)/2.
Calculate
=
2.
in,,
Calculate
1.
—
Let f(x) be a continuous function, and let a
1 and b
1 be numbers satisfying
<
<
1
a
b and f(a
) f(b
1
)
1
0. Let E denote the desired bound for the error
Ir nzI. Repeat steps ito 5 for n 1,2,... until h,, < E:
Y
Second step
‘2
x)
)
1
f(m
Answers to Concepts Review: 1. continuous; (a, b);
f(b) f(a) = f’(c)(b a) 2. f’(O) does not exist
4+ C
3. F(x) = G(x) + C 4. x
Figure 2
y
Figure 1
First step
x
191
Begin the process by sketching the graph of f, which is assumed to be a contin
uous function (see Figure 1). A real root r of f(x) = 0 is a point (technically, the
x-coordinate of a point) where the graph crosses the x-axis. As a first step in pin
ning down this point, locate two points, a
1 < b
, at which you are sure that f has
1
opposite signs; if f has opposite signs at a
, then the product f(a
1
1 and b
) .f(b
1
)
1
will be negative. (Try choosing a
1 and b
1 on opposite sides of your best guess at r.)
The Intermediate Value Theorem guarantees the existence of a root between a
1
and b
1 + b
)/2 of [a
1
. Now evaluate fat the midpoint m
1
1 = (a
,b
1
]. The number
1
1 is our first approximation to r.
nz
Either
= 0, in which case we are done, or f(ni
) differs in sign from
1
). Denote the one of the subintervals [a
1
,m
1
] or [m
1
,b
1
1 on which the
f (ar) orf(b
sign change occurs by the symbol [a
,b
2
], and evaluate f at its midpoint
2
= (a
2 + b
)/2 (Figure 2). The number m, is our second approximation to r.
2
Repeat the process, thus determining a sequence of approximations
, fl12, m
1
m
,..., and subintervals [a
3
,b
3
],.., each subinterval con
3
,b
1
], [a
1
,b
2
], [a
2
taining the root r and each half the length of its predecessor. Stop when r is deter
mined to the desired accuracy, that is, when (b,,
)/2 is less than the allowable
0
a
error, which we will denote by E.
Section 3.7 Solving Equations Numerically
Algorithm
T
[A,B].
54. Show that if an object’s position function is given by
s(t) = at
2 + bt + c, then the average velocity over the interval
[A, B] is equal to the instantaneous velocity at the midpoint of
53. A car is stationary at a toll booth. Twenty minutes later at
a point 20 miles down the road the car is clocked at 60 miles per
hour. Explain why the car must have exceeded 60 miles per hour
at some time after leaving the toll booth, but before the car was
clocked at 60 miles per hour.
52. A car is stationary at a toll booth. Eighteen minutes later
at a point 20 miles down the road the car is clocked at 60 miles
per hour. Sketch a possible graph of v versus t. Sketch a pos
sible graph of the distance traveled s against t. Use the Mean
Value Theorem to show that the car must have exceeded the
60 mile per hour speed limit at some time after leaving the toll
booth, but before the car was clocked at 60 miles per hour.
51. John traveled 112 miles in 2 hours and claimed that he
never exceeded 55 miles per hour. Use the Mean Value Theorem
to disprove John’s claim. Hint: Let fQ) be the distance traveled in
time t.
50. Give an example of a function f that is continuous o
[0, 11, differentiable on (0, 1), and not differentiable on [0, 11, and
has a tangent line at every point of [0, 1].
In mathematics and science, we often need to find the roots (solutions) of
an equation f(x) = 0. To be sure, if f(x) is a linear or quadratic polynomial, for
rnulas for writing exact solutions exist and are well known. But for other alge
braic equations, and certainly for equations involving transcendental functions,
formulas for exact solutions are rarely available. What can be done in such
cases?
There is a general method of solving problems known to all resourceful peo
ple. Given a cup of tea, we add sugar a bit at a time until it tastes just right. Given
a stopper too large for a hole, we whittle it down until it fits. We change the solu
tion a bit at a time, improving the accuracy, until we are satisfied. Mathemati
cians call this the method of successive approximations, or the method of
iterations.
In this section, we present three such methods for solving equations: the
Bisection Method, Newton’s Method, and the Fixed-Point Method. All are de
signed to approximate the real roots of f(x) = 0, and they all require many com
putations. You will want to keep your calculator handy.
—
46. Suppose that in a race, horse A and horse B begin at the
same point and finish in a dead heat. Prove that their speeds were
identical at some instant of the race.
47. In Problem 46, suppose that the two horses crossed the
finish line together at the same speed. Show that they had the
same acceleration at some instant.
48. Use the fvtean Value Theorem to show that the graph of a
concave up function f is always above its tangent line; that is,
show that
.r c
f(x) > f(c) + f’(c)(.r c),
Isinx
45. Use the ?vlean Value Theorem to show that
for all
2 in (a, b). Hint: Apply Problem 41 with
and x
f(x) = h(x) g(x).
44. Use the Vlean Value Theorem to prove that
<
43. Prove that if g(x)
42. Prove that if f(.e)
nondecreasing on I.
190 Chapter 3 Applications of the Derivative
‘1