Math 161

Sample Midterm 2

Problem 3

Dr. Wilson

3. 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 - intercept: let x = 0

y = 0

x - intercept: let y = 0

In order for a fraction to be equal to 0, the top has to be 0.

x = 0

Vertical asymptotes: There will be a vertical asymptote when the bottom is 0. In this case, x2 + 1 > 1, so there are no vertical asymptotes.

Since the degree of the bottom is bigger than the degree of the top, the x - axis is an horizontal asymptote.

To determine where the function is positive and where it is negative, we should plug in a positive and a negative number. We can kill two birds with one stone if use the points where the first derivative is 0.

y' = 0 when the top is 0

1 - x2 = 0

x2 = 1

So we will use x = 1 and x = -1 to tell where the function is positive or negative.

so we conclude that f(x) < 0 if x < 0.

so we conclude that f(x) > 0 if x > 0.

We next determine where the function is increasing and decreasing. For this we will choose points between the roots of the first derivative. The best points would be the roots of the second derivative.

Note that we can factor a factor of (x2 + 1) from both of the terms in the top.

The factor of x2 + 1 which we factor out of the top will cancel with one of the factors of x2 + 1 in the bottom to leave us with

which cleans up to

y" = 0 when the top = 0

2x3 - 6x = 0

Factor

2x(x2 - 3) = 0

Set the factors = 0.

2x = 0

x = 0

x2 - 3 = 0

x2 = 3

so f(x) is decreasing when x < -1,

so f(x) is increasing when -1 < x < 1, and

so f(x) is decreasing when x > 1.

We conclude that since f(x) stops decreasing and starts increasing when x = -1, f(x) has a minimum when x = -1..

Since f(x) stops increasing and starts decreasing when x = 1, f(x) has a max when x = 1.

To check concavity, we plug points in between the roots of the second derivative into the second derivative.

so f(x) is concave down when x = -2.

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

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

We know that if f(x) is positive and has a max at x = 1 and the x - axis is a horizontal asymptote, we know that f(x) has to have a point of inflection to the right of x = 1, because it the function is concave down at x =1 and remains concave down, it will descend more and more steeply after leaving the max and eventually cross the x - axis which is not consistent with the x-axis being a horizontal asymptote. Similarly we know that there will have to be a point of inflection to the left of x = -1.

This gives us the following points that we have plotted

and the graph looks like