Math 161

Sample Midterm 2

Problem 4

Dr. Wilson

4. 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 = 1/0

there is no y - intercept

x - intercept.: let y = 0.

a fraction is 0 when its top is 0. Set

x + 1 = 0

x = -1

Vertical asymptotes happen when the bottom is zero. Set the bottom = 0

x2 - 3x = 0

factor

x(x - 3) = 0

set the factors = 0

x = 0

x - 3 = 0

x = 3

Horizontal asymptotes: since the bottom has bigger degree than the top, the x - axis is an horizontal asymptote.

To find where the function assumes positive values and negative values, we will plug some points in between the roots of the top and bottom. The most interesting points to try will be the places where the first derivative is 0. We first find the first derivative. Since our function is a rational function, we will use the quotient rule to find its derivative.

This cleans up to

When we set y' = 0, a fraction is 0 when its top is 0

-x2 - 2x + 3 = 0

If we multiply both sides by -1, we get something that looks a little more familiar.

x2 + 2x - 3 = 0

This factors

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

Set the factors = 0

x - 1 = 0

x = 1

x + 3 = 0

x = -3

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

= -2/(9 + 9

= -2/18

= -1/9

which is negative. We conclude that

f(x) < 0 when x < -1.

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

= 2/(1 - 3)

= 2/(-2)

= -1

which is also negative. We conclude that

f(x) < 0 when 1 < x < 3.

Is it ever positive. There are two intervals that we haven't checked, the interval from -1, the root of the top, to 0, the left root of the bottom, and the interval of numbers which are greater than 3, the bigger root of the bottom.. The simplest number to check in the interval between -1 and 0 is -1/2

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

= (1/2)/(1/4 + 3/2)

= (1/2)/(7/4)

We invert and multiply to get

= (1/2)(4/7)

= 2/7 > 0

so

f(x) > 0 when -1 < x < 0.

The simplest number which is greater than 4 to plug into our function would be 4.

f(4) = ((4) + 1)/((4)2 - 3(4))

= 3/(16 - 12)

=3/4 > 0

so

f(x) > 0 when x > 3.

To tell where the function is increasing and decreasing, we check the sign of y'. y' is given by a rational function whose bottom is a square. Since squares are never negative, the sign on the first derivative depends on the top. Since the top is a quadratic with negative square term, it is negative for x's to the left of the smaller root, positive for x's between the roots, and negative for x's to the right of the bigger root. We conclude that

f(x) is decreasing for x < -3

f(x) is increasing for -3< x < 1

except when x = 0 when the function doesn't exist, and

f(x) is decreasing for x > 1

except when x = 3 when the function also doesn't exist.

For points of inflection and concavity, we check the second derivative. Since the first derivative is a rational function, we will use the quotient rule.

We will want to clean this up. Let's factor out a 2(x2 - 3x) from both of the terms in the top and distribute the negative in front of the second term through the first factor in the second term.

The factor of x2 - 3x that we factored out of the top will cancel with one of the factors of x2 - 3x in the bottom makingit easier to simplify the top. We will take 2 steps to simplify the top. First we will multiply the polynomials inside the parentheses after the 2 on top.

Now we combine like terms inside the parentheses on top.

We won't bother to multiply the 2 by all the terms in the top, because in order for a fraction to be 0, the top has to be zero. When we set

2(x3 + 3x2 - 9x + 9) = 0

since 2 is not 0, we can divide both sides by 2 and get

x3 + 3x2 - 9x + 9 = 0

This gives us a cubic. Unfortunately, it is a cubic that doesn't factor. Fortunately, there is a cubic formula. We first make the substitution

x = y - 1

Our equation then becomes

(y - 1)3 + 3(y - 1)2 - 9(y - 1) + 9 = 0

rRemoving parentheses

y3 - 3y2 + 3y - 1 + 3y2 - 6y + 3 - 9y + 9 + 9 = 0

Combining like terms

y3 - 12y + 20 = 0

This gives us a depressed cubic as cubics that don't have square terms are called. We can use the formula for the depressed cubic

if y3 + py + q = 0

then

In this case, q = 20 and p = -12, and we get

so

This is the only real point of inflection. With the cubic formula, if the number under the square root under the cube root is positive, there will only be one real root.

If we want to tell where the function is increasing and where it is decreasing by plugging numbers into the first derivative, while there is a tremendous amount of fun to be had by plugging this number into the first derivative, we can tell whether the function is increasing or decreasing over the interval to the left of the smaller root of the derivative just as well by plugging another easier number in the interval into the first derivative such as -5. When x = -5

y' = (-(-5)2 - 2(-5) + 3)/((-5)2 - 3(-5))2

= (-25 + 10 + 3)/(25 + 15)2

= -12/(40)2

= -12/1600

= -3/400

= -0.0075 < 0

and we conclude that

y is decreasing when x < -3

The next interval where the first derivative could change sign would be the interval between x = -1 where y' = 0 and x = 0 where y' doesn't exist. The simplest number in that interval is -1/2. We let x = -1/2 in the first derivative and get

y' = (-(-1/2)2 - 2(-1/2) + 3)/((-1/2)2 - 3(-1/2))2

= (-1/4 + 1 +3)/(1/4 + 3/2)2

= (15/4)/(7/4)2

= (15/4)/(49/16)

We are ready to incert and multiply

= (15/4)(16/49)

= 60/49 > 0

and we conclude that

y is increasing when -1 < x < 0

The next interval is between x = 0 where y' doesn't exist and x = 1 where y' = 0. The simplest number in this interval is x = 1/2

y' = (-(1/2)2 - 2(1/2) + 3)/((1/2)2 - 3(1/2))2

= (-1/4 - 1 +3)/(1/4 - 3/2)2

= (7/4)/(-5/4)2

= (7/4)/(25/16)

We are ready to incert and multiply

= (7/4)(16/25)

= 28/25 > 0

and we conclude that

y is increasing when 0 < x < 1

The next interval is between x = 1 where y' doesn't exist and x = 3 where y' = o. The simplest number in this interval is x = 2

y' = (-(2)2 - 2(2) - 3)/((2)2 - 3(2))2

= -3/(4 - 6)2

= -3/(-2)2

= -3/4 < 0

and we conclude that

y is decreasing when 1 < x < 3

The last interval is the set of x's where x > 3. The simplest such number is 4. When x = 4

y' = (-(4)2 - 2(4) - 3)/((4)2 - 3(4))2

(-16 - 8 - 3)/(16 - 12)2

= -27/(4)2

= -27/16 < 0

and we conclude that

y is decreasing when x > 3.

Thus

y has a min when x = -1 because the function stops decreasing and starts increasing, and y has a max at x = 1 because the function stops increasing and starts decreasing.

Note that the derivative did not change signs when the function went past the roots of the bottom. This is because the bottom is squared in the first derivative. Even though the amount inside the parentheses in the bottom changed sign, it didn't matter because the bottom was being squared.

To check for concavity, we consider the intervals between the points where the second derivative is either 0 or doesn't exist. There was only the one point where the second derivative was 0 close to when x = -5, however, the second derivative also doesn't exist at the vertical asymptotes.

The first intereval to check for concavity then would be to the left of the point where the second derivative was 0. since this was close to -5, we could check this interval by plugging -6 into y". When x = -6

y" = 2((-6)3 + 3(-6)2 - 9(-6) + 9)/((-6)2 - 3(-6))3

= 2((-6)3 + 3(36) - 9(-6) + 9)/(36 - 3(-6))3

= 2(-216 + 108 + 54 + 9)/((36) +18)3

= 2(-45)/(54)3

= - 90/157464

= - 5/8748 < 0

and we conclude that

the function is concave down if x < -5.1072...

The next interval when x is between this point where the second derivative is 0 and x = 0 where the it doesn't exist. The simplest such number to check would be x = -1

y" = 2((-1)3 + 3(-1)2 - 9(-1) + 9)/((-1)2 - 3(-1))3

= 2(-1 + 3 + 9 + 9)/(1 + 3)3

= 2(20)/(4)3

= 40/8

= 5 > 0

and we conclude that

the function is concave up if -5.1072... < x < 0

The next interval when x is between the points where the second derivative doesn't exist, ie the interval where 0 < x < 3. The simplest such number to check would be x = 1

y" = 2((1)3 + 3(1)2 - 9(1) + 9)/((1)2 - 3(1))3

= 2(1 + 3 - 9 + 9)/(1 - 3)3

= 2(4)/(-2)3

= 8/(-8)

= -1 < 0

and we conclude that

the function is concave down if 0 < x < 3

The last interval to check would be the interval where x > 3. The simplest such number to check would be x = 4

y" = 2((4)3 + 3(4)2 - 9(4) + 9)/((4)2 - 3(4))3

= 2(64 + 3(16) - 9(4) + 9)/(16 - 12)3

= 2(64 + 48 - 36 + 9)/(4)3

= 2(75)/64

= 75/32 > 0

and we conclude that

the function is concave up if x > 3

Note that in the second derivative, since the bottom is cubed, there is a sign change when you go across the roots of the bottom.

This gives us the following points to plot

which gives us th following graph