Math 161

Sample Midterm 2

Problem 2

Dr. Wilson

2. Find any x and y intercepts, asymptotes, horizontal, vertical, or otherwise, find where the function assumes positive values and where it assumes negative values, the places where the first derivative is either 0 or does not exists, where the function is increasing and decreasing, find the places where the second derivative is 0 or doesn't exist, and where the function is concave up and concave down. Find the local maxima and minima and points of inflection, and sketch the graph of

y = 3x5 - 5x 3

y - intercept: let x = 0

y = 0

x -ntercepts or roots: let y = 0

0 = 3x5 - 5x 3

 Factor

x3(3x2 - 5) = 0

Set the factors = 0

x3 = 0

x = 0

3x2 - 5 = 0

3x2 = 5

x2 = 5/3

To see where the function is positive and where it is negative we should pick points between the roots, and see what kind of answers we get. We can be a little judicious about which points to pick if we use the places where the first derivative is 0.

f'(x) = 15x4 - 15x2

Set f'(x) = 0

15x4 - 15x2

Factor

15x2(x2 - 1) = 0

Set the factors = 0

15x2 = 0

x2 = 0

x = 0

x2 - 1 = 0

x2 = 1

Since

f(-1) = 3(-1)5 - 5(-1)3 = -3 + 5 = 2 > 0,

we conclude that

and since

f(1) = 3(1)5 - 5(1)3 = 3 - 5 = -2 < 0,

we conclude that

To see whether f(x) is positive or negative if x2 > 5/3, consider x = -2 and 2.

f(-2) = 3(-2)5 - 5(-2)3 = -96 + 40 = -56 < 0,

we conclude that

and since

f(2) = 3(2)5 - 5(2)3 = 96 - 40 = 56 > 0,

we conclude that

To see where f'(x) is increasing or decreasing, we check points between the roots of the first derivative. Let us first check for points of inflection. Set the second derivative = 0.

f"(x) = 60x3 - 30x

Set f"(x) = 0

60x3 - 30x = 0

Factor

30x(2x2 - 1) = 0

Set the factors = 0

30x = 0

x = 0

2x2 - 1 = 0

2x2 = 1

x2 = 1/2

so f(x) is decreasing if -1 < x < 0.

so f(x) is also decreasin if 0 < x < 1.

This is an example of a case where the function does not have a max or min at a place where the first derivative is 0. f(x) is decreasing right before x = 0 and it is decreasing right after x = 0, so there is no max or min when x = 0. The fact that the first derivative is 0 means that the tangent line is horizontal when x = 0, but it is horizontal at just the point where x = 0.

We also need to check to see if it is increasing or decreasing outside of -1 and 1.

f'(-2) = 15((-2)4 - (-2)2) = 15(16 - 4) = 15(12) = 180

so f(x) is increasing when x = -2

f'(2) = 15(24 - 22) = 15(16 - 4) = 15(12) = 180

so f(x) is increasing when x = 2.

We conclude that f(x) is increasing when x < -1, decreasing when -1 < x < 1, and increasing again when x > 1.

As a result, since f(x) stops increasing and starts decreasing when x = -1, f(x) has a relative max when x = -1. Since f(x) stops decreasing and starts increasing when x = 1, f(x) has a relative max when x = 1.

We now check for concavity. We plug numbers between the roots of the second derivative into the second derivative to see whether it is positive or negative at those points.

f"(x) = 60x3 - 30x

so

f"(-1) = 60(-1)3 -30(-1)

= -60 + 30

= -30 < 0

so f(x) is concave down when x = -1.

f"(-1/2) = 60(-1/2)3 -30(-1/2)

= - 15/2 + 15

= 15/2 > 0

so f(x) is concave up when x = -1/2

f"(1/2) = 60(1/2)3 -30(1/2)

= 15/2 - 15

= -15/2 < 0

so f(x) is concave down when x = 1/2

f"(1) = 60(1)3 -30(1)

= 60 - 30

= 30 > 0

so f(x) is concave up when x = 1.

We have plotted the following important points

which gives us the following graph