Week 4: Rational Functions – Answers

Week 4: Rational Functions – Day 5

Here is an organized way to do this.

A. Find the zeros and undefined x-values of the function (places where numerator or denominator are zero).
B. Arrange these values in increasing order.
C. Divide the x-axis into intervals around these values. For example, if the values are -1 and 0, the intervals are:
-1 < x, -1 < x < 0, and 0 < x. If there are N values, there will be N+1 intervals.
D. Test the sign of the function on each interval. You don’t have to compute exact values of the function, just check to see if it is positive or negative.

Here is a set of problems that guide through the process:

1. f(x) = (x2 + x) / (x2 + 6x + 5)
Factor the numerator and denominator of f(x).

f(x) = x(x + 1) / (x + 1)(x + 5)
As long as x does not equal -1, f(x) = x/(x + 5)

2. Where does f(x) have zero value(s)?

f(x) has only one zero value, at x = 0. (It might seem like f(x) is zero at x = -1, but f(x) is undefined there.)

3. Where does f(x) have vertical asymptote(s)?

f(x) has only one vertical asymptote, at x = -5. (It might seem like f(x) has a vertical asymptote at x = -1, but f(x) does not approach infinity or minus infinity near x = -1.)

4. Where does f(x) have a hole?

f(x) has a hole at its undefined point, x = -1. Because f(x) has defined, bounded (not unbounded) values near x = -1, that point is a hole in the graph.

5. Using the information from problems 1-4, test the sign of f(x) over the appropriate intervals.

The values we have to pay attention to are: -5, -1, and 0. The intervals are:
x < -5
-5 < x < -1
-1 < x < 0
0 < x

If x < -5, f(x) is positive.
If – 5 < x < -1, f(x) is negative.
If – 1 < x < 0, f(x) is negative.
If 0 < x, f(x) is positive.


 

Week 4: Rational Functions – Day 4

What is the end behavior of each of these functions? If there is a horizontal or slant asymptote, what is its equation?

1. f(x) = (2x + 1) / (x – 4)

The degree of the numerator and denominator is the same. There is a horizontal asymptote. To find its equation, we divide numerator and denominator by the highest power of x (in effect, multiply f(x) by the x-1 / x-1)

f(x) = (2 + 1/x) / (1 – 4/x)

for large values of x, the 1/x and 4/x have very little effect and f(x) is very close to 2/1 = 2.

Therefore the horizontal asymptote is y = 2.

2. f(x) = (x2 + 3) / (x – 1)

The degree of the numerator is 2, the degree of the denominator is 1. Since 2-1 = 1, there is a slant asymptote.
To find the equation of the slant asymptote, we use long division:

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3. f(x) = (3x4 – x3 + 5) / (x2 + 7)

Degree of numerator is 4, degree of denominator is 2. Since 4-2 = 2 > 1, there is no linear asymptote. The end behavior of this function is similar to 3x4 / x2 = 3x2.

In other words, the end behavior is that of the quadratic 3x2 ; it goes to infinity as x goes to infinity or negative infinity.

4. f(x) = 17 / (x2 + 4x – 5)

The degree of the numerator is zero, and the degree of the denominator is 2. This function tends to zero for large positive or negative values of x. The horizontal asymptote is y=0.

5. f(x) = (6x2 + 1) / (3x2 – 1)

The degree of the numerator is 2, and degree of denominator is 2. Therefore there is a horizontal asymptote.
This function is equivalent to:

f(x) = ( 6 + 1/x2) / (3 – 1/x2)

For large values of x, f(x) is very close to 6/3 = 2. The horizontal asymptote is y = 2.


 

Week 4: Rational Functions – Day 3

Which functions below have vertical asymptotes? If there are any vertical asymptotes, what are their equations?

1. f(x) = 3 / (x – 2)

One vertical asymptote, x = 2.

2. f(x) = (x3 + x)/x

This function is equivalent to x2 + 1 for x not equal to zero. There is no vertical asymptote, even though it is undefined at x = 0.

3. f(x) = x / (x2 + 5)

This function is defined for all x. It does not approach infinity for finite values of x (it has a defined, finite value of y for every finite value of x). There is no vertical asymptote.

4. f(x) = (2x3 – 32)/(x2 – 6x + 5)

This function has a denominator equivalent to (x – 5)(x – 1). So it is undefined at x = 5 and x = 1. The numerator is nonzero at x = 5 and x = 1, and therefore there are two asymptotes, x = 5 and x = 1.

5. f(x) = 1 – 1/x + 2/(x – 1)

Week 4: Rational Functions – Day 2

The range is the set of all values taken by the function. State the range of the following rational functions, in interval notation.

1. f(x) = 1/(x – 1)

This function never takes the value zero because the numerator is 1, and never zero (a quotient is zero only if the numerator is zero). This function reaches negative values if x < 1 and positive values if x > 1. The function approaches negative infinity as x approaches 1 form the left, and positive infinity as x approaches 1 from the right (check this by evaluating f(x) for numbers such as x = 0.9999 and x = 1.0001 ; you will see that f approaches negative and positive infinity).

The range is the set (-∞, 0) U (0, ∞)

2. f(x) = 3

The range is the single value 3.

3. f(x) = 3/(x2 + 1)

This function never takes the value zero because the numerator is 3, which is never zero. This function is always positive. It has a maximum value of 3, achieved when x = 0. It reaches very small values when x is large (plug x = 1000 into f(x) for example).

The range is (0, 3]

4. f(x) = 1/x2

This function never takes the value zero. It is always positive. It approaches infinity for x close to zero (plug x = 0.0001 or x = -0.00001 into f(x) for example). For large values of x, it approaches zero.

The range is (0, ∞)

5. f(x) = (x3 + 1)/x

To see what the range is, it helps to divide through by x.

f(x) = x2 + 1/x

This function has multiple effects happening. The x2 part is positive for all x, but the 1/x part is negative for x

The range is (-∞, ∞).


 

Week 4: Rational Functions – Day 1

Definition, Domain

The definition of a rational function is: it can be written as a quotient of two polynomials.

1. Suppose f(x) = 1/x and g(x) = 1/(x + 1). Are f(x) and g(x) both rational functions?

Yes. f(x) is the quotient of the polynomials 1 and x. g(x) is the quotient of the polynomials 1 and (x + 1).

2. Is h(x) = f(x) + g(x) from problem 1 above a rational function? Explain.

Yes. Finding a common denominator, h(x) = (2x + 1) / (x2 + x) and is therefore the quotient of two polynomials.

3. What is the domain of f(x) = 1/x?

f(x) is only undefined for x = 0. The domain of f(x) is (-∞, 0) U (0, ∞) in interval notation. Put another way the domain of f is the set of x such that x does not equal zero.

4. What is the domain of the rational function f(x) = (x – 1) / (x2 + 1) ?

This function is defined for all x. The domain is (-∞, ∞).

5. What is the domain of h(x) = 1/x + 1/(x + 1)?

h(x) is not defined for x = 0 and x = -1. The domain of h is the set of all x not equal to 0 or -1. In interval notation: (-∞, -1) U (-1, 0) U (0, ∞)