Math 103. Homework 8. Solutions

1. [33, p. 293] Find the intervals where the function

f(x)= x2-1 x

is decreasing and the intervals where it is increasing.

Solution. By the quotient rule, the first derivative is

f¢(x) = 2x(x) -(x2-1) x2 = x2+1 x2

The equation f¢(x)=0 is

x2+1 x2 =0

or

x2+1=0

which has no real solutions.

The derivative is not defined when x=0. In fact, x=0 is not in the domain of f(x).

The sign graph of the derivative is

Therefore, f(x) is increasing on (-¥,0)È(0,¥).

2. Find the intervals where the function

f(x) = 2 1+x2

is concave upward and the intervals where it is concave downward.

Solution. We need the second derivative. The first is

f¢(x) = -2(2x) (1+x2)2

and the second

f¢¢(x) = -4(1+x2)2-(-4x)(2)(1+x2)(2x) (1+x2)4 = -4(1+x2)2+16x2(1+x2)2 (1+x2)4

Setting f¢¢(x)=0 yields

-4(1+x2)2+16x2(1+x2)2=0

and factoring out (1+x²) gives

(1+x2)(-4(1+x2) +16x2)=0

Since 1+x² is never 0, we are left with

-4(1+x2) +16x2 = -4-4x2 +16x2 = 0

or

x=±   æ  ú Ö 1 3

The second derivative exists at all real numbers because the denominator is (1+x²)⁴ which is never 0.

The sign graph for f¢¢(x) is

Therefore, f(x) is concave up on (-¥,-(1/3)^1/2)È ((1/3)^1/2,¥), and concave down on (-(1/3)^1/2,(1/3)^1/2)

3. Exercises 44-45-46-47, p. 294. (These 4 exercises form part of a single one.)

Solution. 44-(c), 45-(a), 46-(b), 47-(d).

4. [58, p. 294] Find the relative maxima and relative minima, if any, of the function f(x)=x⁴-6x²+8x-16.

Solution. The first derivative

f¢(x) = 4x3 -12 x +8

Setting f¢(x)=0 yields

4x3 -12 x +8 = 0

and the solutions are

x = 1, -2

(x=1 is a double root).

The sign graph for the derivative is

Therefore f reaches a relative minimum at x=-2. The relative minimum value is f(-2) = (-2)⁴-6(-2)² +8(-2)-16 = -40.

5. The total annual revenue R of the Miramar Resorts Hotel is related to the amount of money x the hotel spends in advertising its services by the function

R(x) = -0.003 x3 +1.35 x2 +2x +8000       (0 £ x £ 400)

where both R and x are measured in thousands of dollars. Find the inflection point of R and discuss its significance.

Solution. An inflection point is a point on the graph where there is a change in concavity. We thus need the second derivative R¢¢(x).

The first is R¢(x) = -0.009 x² +2.7 x +2, and the second is

R¢¢(x) = -0.018 x +2.7

Setting -0.018 x +2.7=0 gives x=150. The sign graph of the second derivative is

We see that R changes from concave up to concave down as we move across x=150. Therefore, the point (150, 28550) is the inflection point of the graph of R(x).

The revenue is increasing at an increasing rate until the amount of money spent on advertising is $100,000. After that, the revenue is still increasing but at a decreasing rate.