Math 161

Sample Midterm 2

Problem 1

Dr. Wilson

1. y = x3 - 3x

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.

y - intercept. Let x = 0.

y = (0)3 - 3(0) = 0

x - intercept. Let y = 0.

0 = x3 - 3x

We now need to solve for x. To solve a polynomial equation, transpose all terms to one side, leaving a 0 on the other and hope that it factors. Not only do we already have a 0 on the right side, it factors.

0 = x(x2 - 3)

Set the factors = 0.

x = 0

x2 - 3 = 0

x2 = 3

This gives us all the roots or values of x for which f(x) = 0. Since the function is a polynomial and polynomials are continuous functions, the x's for which f(x) = 0 will separate the x's for which f(x) > 0 from the x's for which f(x) < 0. To find where f(x) > 0 and f(x) < 0, will need to evaluate the function at points in between these roots. The points which will be most advantageous to check will be points where f'(x) = 0 or doesn't exist. We find the first derivative

y' =3x2 - 3

This always exists. The extreme points will then be where the first derivative is 0. We set

3x2 - 3 = 0

This factors

3(x2 - 1) = 0

3(x + 1)(x - 1) = 0

Since 3 is never 0, we can divide both sides by 3 to get

(x + 1)(x - 1) = 0

Set the factors = 0

x + 1 = 0

x = -1

x - 1 = 0

x = 1

Since

f(-1) = (-1)3 - 3(-1) = -1 + 3 = 2,

we conclude that

Since

f(1) = (1)3 - 3(1) = 1 - 3 = 2

we conclude that

We still need to know what is happening if x2 > 3. The simplest such x's to check would be 2 and -2.

f(-2) = (-2)3 - 3(-2) = -8 + 6 = -2

so

and

f(2) = (2)3 - 3(2) = 8 - 6 = 2

so

Since f(x) is increasing when f'(x) > 0 and decreasing when f'(x) < 0, we use these same techniques on f' to tell when the function is increasing and decreasing. We know when f'(x) = 0, at 1 and -1. We break the real number line into the intevals between these roots. To see whether f'(x) is positive or negative when x < -1, we check and see what happens when x = -2.

f''(-2) = 3((-2)2 -1) = 3(4 - 1) = 3(3) = 9 > 0

so

f'(x) > 0 and f(x) is increasing if x < -1.

To see what is happening between -1 and 1, we check out x = 0

f'(0) = 3((0)2 - 1) = 3(-1) = -3 < 0

so

f'(x) < 0 and f(x) is decreasing when -1 < x < 1.

To see what is happening if x > 1, we check what happens if x = 2

f'(2) = 3((2)2 - 1) = 3(4 - 1) = 3(3) = 9 > 0

so

f'(x) > 0 and f(x) is increasing if x > 1.

At this point we can classify the critical points 1 and -1. f has a relative max when x = -1, because the function stops increasing and starts decreasing. f has a relative min when x = 1 because the function stops decreasing and starts increasing.

Finally we consider concavity and look for points of inflection. For this we need the second derivative.

f"(x) = 6x

f"(x) = 0 if x = 0

We quickly and easily see that f"(x) < 0 if x < 0 and f"(x) > 0 if x > 0. Thus f(x) is concave down when x < 0 and concave up when x > 0. Hence there is a point of inflection when x = 0.

We are now ready to sketch the graph. We have plotted the following points

and we can draw the graph.

Polynomials do not have any kind of asymptotes.