Math 1A: Homework 7 Solutions

Math 1A: Homework 7 Solutions

August 6

1. Show that the given equations have exactly one solution in the given interval:

(a) x3 + 4x - 13 2 = 0 in (-4, 10).

Let f (x) = x3 + 4x - 13 2. We first show that f (x) = 0 has at least one

solution in the given interval. Note that f is continuous on [-4, 10] and that f (-4) = -80 - 13 2 < 0 while f (10) = 1040 - 13 2 > 0. By the intermediate

value theorem, it follows that f (c) = 0 for some c in (-4, 10).

Next, we show that f (x) = 0 has at most one solution in the given interval.

Assume that it has more than one solution in (-4, 10); pick two such solutions,

say x = a and x = b, and suppose a < b. Since f is continuous on [a, b],

differentiable on (a, b) and f (a) = f (b) = 0, it follows from Rolle's theorem that f (d) = 0 for some d in (a, b). Note however that f (x) = 3x2 + 4 > 0 for all x so

f (d) cannot be zero. We therefore have a contradiction, showing that f (x) = 0

has at most one solution in the given interval. Couple this with the fact that

f (x) = 0 has at least one solution to conclude that this equation has exactly one

solution in the given interval.

(b) x + sin2

x 3

= 8 in (-, ).

Let f (x) = x + sin2

x 3

- 8. We first show that f (x) = 0 has at least one solution

in the given interval. Note that f is continuous on [0, 20] and that f (0) = -8 < 0

while f (20) = 12 + sin( 4) > 0. By the intermediate value theorem, it follows that

f (c) = 0 for some c in (0, 20) and hence in (-, ).

Next, we show that f (x) = 0 has at most one solution in the (-, ). Assume

that it has more than one solution; pick two such solutions, say x = a and x = b,

and suppose a < b. Since f is continuous on [a, b], differentiable on (a, b) and

f (a) = f (b) = 0, it follows from Rolle's theorem that f (d) = 0 for some d in

(a, b).

Note

however

that

f

(x)

=

1

+

(

2 3

)

sin(x/3)

cos(x/3)

=

1+

sin(2x/3) 3

>

0

for

all

x

since

sin(2x/3) 3

-1/3.

This shows

that

f

(d) cannot be zero.

We therefore

have a contradiction, showing that f (x) = 0 has at most one solution in the

(-, ). Couple this with the fact that f (x) = 0 has at least one solution to

conclude that this equation has exactly one solution in the given interval.

(c)

sec() -

1

+2

=0

in

(0, /2).

Let

f ()

=

sec()

-

1

+

2.

We first show that f () = 0 has at least one so-

lution in the given interval. Note that f is continuous on [/6, /3] and that

1

f (/12)

=

-0.784

<

0

while

f (/3)

=

2-

3

+2

=

4-

3

>

0.

By

the

intermedi-

ate value theorem, it follows that f (c) = 0 for some c in (/12, /3) and hence in

(0, /2).

Next, we show that f (x) = 0 has at most one solution in the (0, /2). Assume

that it has more than one solution; pick two such solutions, say = a and = b,

and suppose a < b. Since f is continuous on [a, b], differentiable on (a, b) and

f (a) = f (b) = 0, it follows from Rolle's theorem that f (d) = 0 for some d in

(a, b).

Note

however

that

f

()

=

sec() tan()

+

1 2

> 0 for all since sec(),

tan()

and

1 2

are

positive

in

(0, /2).

This

shows

that

f

(d)

cannot

be

zero.

We

therefore have a contradiction, showing that f () = 0 has at most one solution in

the (0, /2). Couple this with the fact that f () = 0 has at least one solution to

conclude that this equation has exactly one solution in the given interval.

(d) x3 + ex = 0 in (-, ).

Let f (x) = x3 + ex. We first show that f (x) = 0 has at least one solution in the given interval. Note that f is continuous on [-2, 0] and that f (-2) = -8+e-2 < 0 while f (0) = e0 > 0. By the intermediate value theorem, it follows that f (c) = 0 for some c in (-2, 0) and hence in (-, ). Next, we show that f (x) = 0 has at most one solution in the (-, ). Assume that it has more than one solution; pick two such solutions, say x = a and x = b, and suppose a < b. Since f is continuous on [a, b], differentiable on (a, b) and f (a) = f (b) = 0, it follows from Rolle's theorem that f (d) = 0 for some d in (a, b). Note however that f (x) = 3x2 + ex > 0 for all x since x2 0 and ex > 0. This shows that f (d) cannot be zero. We therefore have a contradiction, showing that f (x) = 0 has at most one solution in the (-, ). Couple this with the fact that f (x) = 0 has at least one solution to conclude that this equation has exactly one solution in the given interval.

2. Prove the following inequalities.

(a) | cos(x) - cos(y)| |x - y|.

Observe first that the inequality holds trivially if x = y. Suppose next that x < y and let f (t) = cos(t). Then, f is continuous on [x, y] and differentiable on (x, y). By the mean value theorem, there exists some c in (x, y) such that

f (y) - f (x) = f (c)

y-x cos(y) - cos(x)

= - sin(c) y-x cos(y) - cos(x)

= | sin(c)| 1 y-x | cos(x) - cos(y)| |x - y|.

2

(b)

1 2 n+1

<

n+

1- n

<

1 2n

whenever n > 0.

Let f (x) = x and fix n > 0. Observe then that f is continuous on [n, n + 1] and differentiable on (n, n + 1). By the mean value theorem, there exists some c in (n, n + 1) such that

f (n + 1) - f (n) = f (c)

n+1-n

1

n+1- n = .

(1)

2c

Note next that

n<

c

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