Written Homework 6 Solutions - University of Texas at Austin

Written Homework 6 Solutions

Section 3.10

30. Explain in terms of linear approximations or differentials why the approximation is reasonable: (1.01)6 1.06

Solution: First start by finding the linear approximation of f (x) = x6 about 1. L(x) = f (a) + f (a) ? (x - a) = f (1) + f (1) ? (x - 1) = (1)6 + 6(1)5(x - 1) = 1 + 6(x - 1)

Now to plug in the value to approximate, 1.01. L(1.01) = 1 + 6(1.01 - 1) = 1 + 6(0.01) = 1.06

The approximation is reasonable because the linear approximation is a good approximation as long as x is reasonably small. In this case x = 0.1.

Homework 6 Solutions

Section 4.1

54.

Find the absolute maximum and absolute minimum values of

f (x) =

x x2-x+1

on the interval [0,3]

Solution: First find the value at the endpoints:

f

(0)

=

0

-

0 0

+

1

=

0

f (3)

=

9

-

3 3

+

1

=

3 7

Just to be sure the function does not diverge on the interval check where the roots of the denominator are.

But note the following:

Ax2 + Bx + C = 1 ? x2 + (-1) ? x + 1 B2 - 4AC = 1 - 4 ? 1 ? 1 = -3 < 0

So the roots are imaginary and the function does not diverge on the interval. Now to find the critical points

of the function.

f

(x)

=

(x2

- x + 1) - x(2x (x2 - x + 1)2

- 1)

=

-x2 + 1 (x2 - x + 1)2

It is clear that f (x) = 0 for x = ?1. Since -1 is not in the domain we can throw this value out. Calculate the

value of f at the critical points:

f

(1)

=

1

-

1 1

+

1

=

1

We now compare all of the values we have calculated (from endpoints and critical points):

Absolute max is 1, occurs at x = 1.

Absolute min is 0, occurs at x = 0.

56. Find the absolute maximum and absolute minimum values of f (t) = 3 t(8 - t) on the interval [0,8]

Solution: Begin by checking the value of the endpoints. f (0) = 3 0(8 - 0) = 0

f (8) = 3 8(8 - 8) = 2 ? 0 = 0

Now to find the critical points. Find the roots of the derivative.

f (t) = t-2/3 ? (8 - t) - t1/3 3

t-2/3 ? (8 - t) - t1/3 = 0 t-2/3 ? (8 - t) = t1/3 (8 - t) = 3t t = 2

3

3

Test this value

f (2) = 3 2(8 - 2) = 6 3 2 > 0.

We now compare all of the values we have calculated (from endpoints and critical points):

Absolute max is 6 3 2, occurs at t = 2. Absolute min is 0, occurs at x = 0 and x = 8.

Page 2 of 5

Homework 6 Solutions

63. If a and b are positive numbers, find the maximum values of f (x) = xa(1 - x)b, 0 x 1.

Solution: Begin by checking the endpoints.

f (0) = 0a(1 - 0)b = 0

Now to find the critical points.

f (1) = 1a(1 - 1)b = 1a(0)b = 0

f (x) = axa-1(1 - x)b + xa ? b(1 - x)b-1 ? (-1) = axa-1(1 - x)b - bxa(1 - x)b-1 = xa-1(1 - x)b-1(a(1 - x) - bx) = xa-1(1 - x)b-1(a - (a + b)x)

Now, f (x) is defined for all x in (0, 1). f (0) will be zero if a > 1, and similarly f (1) will be zero if b > 1.

But we already checked the values at the endpoints, so we only need to look for critical points in the interior

of

the

domain

(0, 1).

The

only zero

of

f

(x)

in

(0, 1)

is

at

x

=

a a+b

.

Now plug in that value.

f

a b+a

=

aa b+a

1-

a b+a

b

=

aa b+a

b b+a

b

=

(a

aabb + b)(a+b)

.

Because a and b are positive and nonzero, this is the maximum.

Answer:

Maximum

of

aa bb (a+b)(a+b)

occurs

at

x

=

a a+b

.

Page 3 of 5

Homework 6 Solutions

Problem 4.2

18. Show that the equation has exactly one real root: x3 + ex = 0.

Solution: Let f (x) = x3 + ex. Note that f (x) is continuous for all x. First use the Intermediate Value Theorem to show that a root does exist. For the problem in question let a = -1 and b = 0. Note that:

(-1)3 + e-1 = -1 + 1 < 0 e

03 + e0 = 1 > 0

So then by the Intermediate Value Theorem there exists a value c [-1, 0] such that f (c) = 0. As we know,

this guarantees that at least one such c to exist but there could be more. Assume there were two roots c1 and c2 such that c1 < c2, f (c1) = f (c2) = 0. Now we apply Rolle's theorem on the interval [c1, c2]. Note that:

f (x) = 3x2 + ex

for all x [c1, c2]

Hence f (x) is continuous and defined on the interval (c1, c2). By assumption, f (c1) = 0 = f (c2). So all of the conditions for Rolle's theorem are met. Thus, there is a number d (c1, c2) such that f (d) = 0. However,

f (x) = 3x2 + ex > 0

for all x.

In particular f (d) > 0. This is a contradiction so the assumption that two roots exist is wrong and there must be exactly one root.

24. Suppose that 3 f (x) 5 for all values of x. Show that 18 f (8) - f (2) 30.

Solution: Use the Mean Value Theorem. The problem states that f (x) exists and is between two values for

all x. Therefore the function f (x) is continuous and differentiable everywhere, particularly on [2, 8]. Then

there exists c such that:

f (c) = f (8) - f (2) 8-2

However 3 f (x) 5 for all x. In particular, 3 f (c) 5.

3 f (c) 5 = 3 f (8) - f (2) 5 = 18 f (8) - f (2) 30 6

This proves the required statement.

26. Suppose that f and g are continuous on [a, b] amd differentiable on (a, b). Suppose also that f (a) = g(a) and f (x) < g (x) for a < x < b. Prove that f (b) < g(b).

Solution: Using the hint in the text look at the function h(x) = f (x) - g(x). Note if h(b) < 0 then the desired result follows. Now apply the Mean Value Theorem to h. Since f and g are continuous on [a, b] and differentiable on (a, b) then so is h (the derivative is linear and the difference of continuous functions is continuous). The conditions of the Mean Value Theorem are statisfied for h so then there exists a c [a, b]

such that: h(b) - h(a) = h (c) ? (b - a)

However recall that

h(a) = f (a) - g(a) = 0

since f (a) = g(a). Also note that

h (c) = f (c) - g (c) < 0

because, by assumption f (x) < g (x) for all x. Finally, because b > a, we have that b - a > 0. Then:

h(b) = h(a) + h (c) ? (b - a) = 0 + h (c) ? (b - a) < 0.

Thus f (b) - g(b) < 0, so f (b) < g(b).

Page 4 of 5

Homework 6 Solutions

Problem A

Let f (x) be differentiable at x = a. The linearization of f at x = a is given by L(x) = f (a) + f (a)(x - a). In class

we said that a good linear approximation should have the property that f (x) - L(x) 0 faster than x - a 0.

Show that this is true. That is, show that

lim f (x) - L(x) = 0 xa x - a

Solution: Recall that the linear approximation about a is given by:

L(x) = f (a) - f (a)(x - a)

Then:

lim f (x) - L(x) = lim f (x) - ( f (a) + f (a)(x - a))

xa x - a

xa

x-a

= lim f (x) - f (a) - f (a)(x - a)

xa

x-a

= lim

f (x) - f (a) f (a)(x - a) -

xa x - a

x-a

= lim

f (x) - f (a) - f (a)

xa x - a

= lim f (x) - f (a) - f (a) xa x - a

= f (a) - f (a)

= 0.

Page 5 of 5

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download