Rational Functions and Limits - Functions - REVIEW OF MAJOR TOPICS - Barron's SAT Subject Test Math Level 2

Barron's SAT Subject Test Math Level 2, 10th Edition (2012)

Part 2. REVIEW OF MAJOR TOPICS

Chapter 1. Functions

1.5 Rational Functions and Limits

The function f is a rational function if and only if f(x) = pg37, where p(x) and q(x) are both polynomial functions and q(x) is not zero. As a general rule, the graphs of rational functions are not continuous (i.e., they have holes, or sections of the graphs are separated from other sections by asymptotes). A point of discontinuity occurs at any value of x that would cause q(x) to become zero.

If p(x) and q(x) can be factored so that f(x) can be reduced, removing the factors that caused the discontinuities, the graph will contain only holes. If the factors that caused the discontinuities cannot be removed, asymptotes will occur.

TIP tip

If p and q are both 0 for some x, there is a hole at that x, but if q is 0 and p is not 0, an asymptote results.

EXAMPLES

1. Sketch the graph of f(x) = pg38.

There is a discontinuity at x = –1 since this value would cause division by zero. The fraction pg39 = (x – 1), and so the graph of f(x) is the same as the graph of y = x – 1 except for a hole at x = – 1.

SAT_SUBJECT_TEST_MATH_LEVEL_2_9TH_ED_0104_001

You could also enter this function on the graphing calculator and get the graph shown above, but it is unlikely that you would see the hole at (–1,0).

TIP tip

Don’t count on seeing holes on a graphing calculator.

2. Sketch the graph of f(x) = pg40.

Since this fraction cannot be reduced, and x = 2 would cause division by zero, a vertical asymptote occurs when x = 2. This is true because, as x approaches very close to 2, f(x) gets either extremely large or extremely small. As xbecomes extremely large or extremely small, f(x) gets closer and closer to zero. This means that a horizontal asymptote occurs when y = 0. Plotting a few points indicates that the graph looks like the figure below.

SAT_SUBJECT_TEST_MATH_LEVEL_2_9TH_ED_0104_002

Vertical and horizontal asymptotes can also be described using limit notation. In this example you could write

  pg41 to mean f(x) gets closer to 0 as x gets arbitrarily large or small

  pg42 to mean f(x) gets arbitrarily large as x approaches 2 from the right

  pg43 to mean f(x) gets arbitrarily small as x approaches 2 from the left

If you entered this function into a TI-83 calculator, your graph will show what appears to be an asymptote. (Actually, the graphing calculator connects the pixel of the largest x-coordinate for which the y-coordinate is negative to the pixel of the smallest x-coordinate for which the y-pixel is positive.) The TI-84 calculators do not connect these two pixels. In either case, you can “read” the graph to determine the two infinite limits.

3. What does pg44 equal?

Since pg45 reduces to x – 1,

pg46

Although the window settings on your calculator may not make it possible to see the hole at x = 1, you can determine the limit of this function as x approaches 1 from both sides by using the table feature. Select Ask for Indpnt; go to TABLE; enter values of x that get progressively closer to 1 from below (e.g., 0.9, 0.99, 0.999, etc.) and above (e.g., 1.1, 1.01, 1.001, etc.); and watch y get closer to 1.

4. What does pg47 equal?

Since “problems” occur only when division by zero appears imminent, this example is extremely easy. As x gets closer and closer to 2, 3x + 5 seems to be approaching closer and closer to 11. Therefore, pg48.

5. What does pg49 equal?

The numerator is always positive, so the graph of this rational function has a vertical asymptote. As x approaches 2 from the right (i.e., 2.1, 2.01, 2.001, . . .), the denominator approaches zero from the right so pg50 gets larger and larger and approaches positive infinity. As x approaches 2 from the left (i.e., 1.9, 1.99, 1.999, . . .), the denominator approaches zero from the left so pg50 gets smaller and smaller and approaches negative infinity. Thus, pg51does not exist since pg52 and pg53.

As in the previous example, you could enter this function into your graphing calculator. Then, with an appropriate window, you could determine the infinite limits as x approaches 2 from the left and right.

6. If pg54

As x approaches zero, 3x + 2 approaches 2, in spite of the fact that f(x) = 0 when x = 0. Therefore, pg55.

7. What does pg56 equal?

As x gets larger and larger, the x2 terms in the numerator and denominator “dominate” in the sense that the terms of lower degree become negligible. Therefore, the larger x gets, the more the rational function looks like pg57.

You can also see this by dividing each term of the numerator and denominator by x2, the highest power of x.

pg58

Now, as pg59, each approaches zero. Thus, the entire fraction approaches pg60 . Therefore, pg61.

To use a graphing calculator to find this limit, enter the function, and use the TABLE in Ask mode to enter larger and larger x values. The table will show y values closer and closer to 1.5.

EXERCISES

1. To be continuous at x = 1, the value of pg62 must be defined to be equal to

  (A) –1

  (B) 0

  (C) 1

  (D) pg63

  (E) 4

2. If pg64, what must the value of k be equal to in order for f(x) to be a continuous function?

  (A) pg65

  (B) pg65_1

  (C) 0

  (D) 2

  (E) No value of k can make f(x) a continuous function.

3. pg69_1

  (A) 0

  (B) pg70

  (C) pg71

  (D) pg72

  (E) This expression is undefined.

4. pg73

  (A) pg74

  (B) 0

  (C) pg75

  (D) pg76

  (E) pg77

5. Which of the following is the equation of an asymptote of pg78?

  (A) pg79

  (B) x = 1

  (C) pg80

  (D) pg81

  (E) y = 1

Answers and Explanations

Rational Functions and Limits

All of these exercises can be completed with the aid of a graphing calculator as described in the example.

1. (D) Factor and reduce: pg82 . Substitute 1 for x and the fraction equals pg83.

2. (D) Factor and reduce the fraction, which becomes 3x + 2. As x approaches zero, this approaches 2.

3. (B) Factor and reduce pg84 . Substitute 2 for x and the fraction equals pg85.

4. (D) Divide numerator and denominator through by x2. As x pg86, the fraction approaches pg87.

5. (D) Factor and reduce pg88 . Therefore a vertical asymptote occurs when 3x – 1 = 0 or pg89, but this is not an answer choice. As x pg86, pg90 . Therefore, pg91 is the correct answer choice.