Miscellaneous Functions - 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

1.6 Miscellaneous Functions

PARAMETRIC EQUATIONS

At times, it is convenient to express a relationship between x and y in terms of a third variable, usually denoted by a parameter t. For example, parametric equations x = x(t), y = y(t) can be used to locate a particle on the plane at various times t.

EXAMPLES

1. Graph the parametric equations k

Select MODE on your graphing calculator, and select PAR. Enter 3t + 4 into X1T and t – 5 into Y1T. The standard window uses 0 for Tmin and 6.28… (2π) for Tmax along with the usual ranges for x and y. The choice of 0 for Tmin reflects the interpretation of t as “time.” With the standard window, the graph looks like the figure below.

SAT_SUBJECT_TEST_MATH_LEVEL_2_9TH_ED_0109_002

If you use TRACE, the cursor will begin at t = 0, where (x,y) = (4,–5). As t increases from 0, the graph traces out a line that ascends as it moves right.

It may be possible to eliminate the parameter and to rewrite the equation in familiar xy-form. Just remember that the resulting equation may consist of points not on the graph of the original set of equations.

2. Eliminate the parameter and sketch the graph k1

Substituting x for t2 in the second equation results in y = 3x + 1, which is the equation of a line with a slope of 3 and a y-intercept of 1. However, the original parametric equations indicate that x pg93 0 and y pg93 1 since t2 cannot be negative. Thus, the proper way to indicate this set of points without the parameter is as follows: y = 3x + 1 and x pg93 0. The graph is the ray indicated in the figure.

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3. Sketch the graph of the parametric equations pg94

Replace the parameter hk2_oo with t, and enter the pair of equations. The graph has the shape of an ellipse, elongated horizontally, as shown in this diagram.

SAT_SUBJECT_TEST_MATH_LEVEL_2_9TH_ED_0110_002

It is possible to eliminate the parameter, hk2_oo, by dividing the first equation by 4 and the second equation by 3, squaring each, and then adding the equations together.

SAT_SUBJECT_TEST_MATH_LEVEL_2_9TH_ED_0110_003

Here, pg96=1 is the equation of an ellipse with its center at the origin, a = 4, and b = 3 (see Coordinate Geometry). Since –1 pg95 cos 299 pg95 1 and –1 pg95 sin 299 pg95 1, –4 pg95 x pg95 4 and –3 pg95 y pg95 3 from the two parametric equations. In this case the parametric equations do not limit the graph obtained by removing the parameter.

EXERCISES

1. In the graph of the parametric equations pg97

  (A) x pg93 0

  (B) pg98

  (C) x is any real number

  (D) x pg93 –1

  (E) x pg95 1

2. The graph of pg99 is a

  (A) straight line

  (B) line segment

  (C) parabola

  (D) portion of a parabola

  (E) semicircle

3. Which of the following is (are) a pair of parametric equations that represent a circle?

  I. pg100

  II. pg101

  III. pg102

  (A) only I

  (B) only II

  (C) only III

  (D) only II and III

  (E) I, II, and III

PIECEWISE FUNCTIONS

Piecewise functions are defined by different equations on different parts of their domain. These functions are useful in modeling behavior that exhibits more than one pattern.

EXAMPLES

1. Graph the functiono1

You can graph this on your graphing calculator by using the 2nd TEST command to enter the symbols < and ≥. Enter (3 – x2)(x < 1) + (x3 – 4x)(x ≥ 1) into Y1. For values of x less than 1, (x – 1) = 1 and (x ≥ 1) = 0, so for these values only 3 – x2 will be graphed. The reverse is true for values of x greater than 1, so only x3 – 4x will be graphed. This graph is shown on the standard grid in the figure below.

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Absolute value functions are a special type of piecewise functions. The absolute value function is defined as o2

The general absolute value function has the form f(x) = a|x h| + k, with the vertex at (h,k) and shaped like v if a > 0 and like ^ if a < 0. The vertex separates the two branches of the graph; h delineates the domain of all real numbers into two parts. The magnitude of a determines how spread out the two branches are. Larger values of a correspond to graphs that are more spread out.

The absolute value command is in the MATH/NUM menu of your graphing calculator. You can readily solve absolute value equations or inequalities by finding points of intersection.

2. If |x – 3| = 2, find x.

Enter |x – 3| into Y1 and 2 into Y2. As seen in the figure below, the points of intersection are at x = 5 and x = 1.

This is also easy to see algebraically. If |x – 3| = 2, then x – 3 = 2 or x – 3 = –2. Solving these equations yields the same solutions: 5 or 1. This equation also has a coordinate geometry solution: |a b| is the distance between a and b. Thus |x – 3| = 2 has the interpretation that x is 2 units from 3. Therefore, x must be 5 or 1.

3. Find all values of x for which |2x + 3|5.

The graphical solution is shown below.

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The desired values of x are on, or right and left of, the points of intersection: x ≥ 1 or x o3 –4. By writing the inequality as o4, we can also interpret the solutions to the inequality as those points that are more than o5units from o6.

4. If the graph of f(x) is shown below, sketch the graph of (A) | f(x)| (B) f(|x|).

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(A) Since |f(x)| ≥ 0, by the definition of absolute value, the graph cannot have any points below the x-axis. If f(x) < 0, then |f(x)| = –f(x). Thus, all points below the x-axis are reflected about the x-axis, and all points above the x-axis remain unchanged.

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(B) Since the absolute value of x is taken before the function value is found, and since |x| = –x when x < 0, any negative value of x will graph the same y-values as the corresponding positive values of x. Thus, the graph to the left of the y-axis will be a reflection of the graph to the right of the y-axis.

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5. If f(x) = |x + 1| – 1, what is the minimum value of f(x)?

Since |x + 1| ≥ 0, its smallest value is 0. Therefore, the smallest value of f(x) is 0 – 1 = –1. The graph of f(x) is indicated below.

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Step functions are another special type of piecewise function. These functions are constant over different parts of their domains so that their graphs consist of horizontal segments. The greatest integer function, denoted by [x], is an example of a step function. If x is an integer, then [x] = x. If x is not an integer, then [x] is the largest integer less than x.

The greatest integer function is in the MATH/NUM menu as int on TI-83/84 calculators.

6. Five examples of the greatest integer function integer notation are:

  (1) [3.2] = 3

  (2) [1.999] = 1

  (3) [5] = 5

  (4) [–3.12] = –4

  (5) [–0.123] = – 1.

7. Sketch the graph of f(x) = [x].

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TIP tip

On the graphing calculator, you can’t tell which side of each horizontal segment has the open and closed point.

8. What is the range of o7.

Enter int(int(x)/x) into Y1 and choose Auto for both Indpnt and Depend in TBLSET. Set TblStart to 0 and j2Tbl to 0.1. Inspection of TABLE shows only 0s and 1s as Y1, so the range is the two-point set {0,1}.

EXERCISES

1. |2x – 1| = 4x + 5 has how many numbers in its solution set?

  (A) 0

  (B) 1

  (C) 2

  (D) an infinite number

  (E) none of the above

2. Which of the following is equivalent to 1 o3 |x – 2| o3 4?

  (A) 3 o3 x o3 6

  (B) x o3 1 or x ≥ 3

  (C) 1 o3 x o3 3

  (D) x o3 –2 or x ≥ 6

  (E) –2 o3 x o3 1 or 3 o3 x o3 6

3. The area bound by the relation |x| + |y| = 2 is

  (A) 8

  (B) 1

  (C) 2

  (D) 4

  (E) There is no finite area.

4. Given a function, f(x), such that f(x) = f(|x|). Which one of the following could be the graph of f(x)?

  (A) SAT_SUBJECT_TEST_MATH_LEVEL_2_9TH_ED_0115_001

  (B) SAT_SUBJECT_TEST_MATH_LEVEL_2_9TH_ED_0115_002

  (C) SAT_SUBJECT_TEST_MATH_LEVEL_2_9TH_ED_0115_003

  (D) SAT_SUBJECT_TEST_MATH_LEVEL_2_9TH_ED_0115_004

  (E) SAT_SUBJECT_TEST_MATH_LEVEL_2_9TH_ED_0115_005

5. The figure shows the graph of which one of the following?

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  (A) y = 2x – |x|

  (B) y = |x – 1| + x

  (C) y = |2x – 1|

  (D) y = |x + 1| – x

  (E) y = 2|x| – |x|

6. The postal rate for first-class mail is 44 cents for the first ounce or portion thereof and 17 cents for each additional ounce or portion thereof up to 3.5 ounces. The cost of a 3.5-ounce letter is 95¢. A formula for the cost in cents of first-class postage for a letter weighing N ounces (N o3 3.5) is

  (A) 44 + [N – 1] · 17

  (B) [N – 44] · 17

  (C) 44 + [N] · 17

  (D) 1 + [N] · 17

  (E) none of the above

7. If f(x) = i, where i is an integer such that i o3 x < i + 1, the range of f(x) is

  (A) the set of all real numbers

  (B) the set of all positive integers

  (C) the set of all integers

  (D) the set of all negative integers

  (E) the set of all nonnegative real numbers

8. If f(x) = [2x] – 4x with domain 0 o3 x o3 2, then f(x) can also be written as

  (A) 2x

  (B) –x

  (C) –2x

  (D) x2 – 4x

  (E) none of the above

Answers and Explanations

Parametric Equations

* 1. (B) Graph these parametric equations for values of t between –5 and 5 and for x and y between –2.5 and 2.5.

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  Apparently the x values are always greater than some value. Use the TRACE function to move the cursor as far left on the graph as it will go. This leads to a (correct) guess of o9. This can be verified by completing the square on the x equation:
o10.

  This represents a parabola that opens up with vertex at o11. Therefore, o12.

2. (D) D is the only reasonable answer choice. To verify this, note that o13. So o14. Adding this to x = sin2t gives k2. Since 0 o3 x o3 1 because 0 o3 sin2 t o3 1, this can only be a portion of the parabola given by the equation y2 + 4x = 4.

* 3. (A) You could graph all three parametric pairs to discover that only I gives a circle. (II and III give semicircles). You can also see this by a simple analysis of the equations. Removing the parameter in I by squaring and adding gives x2 + y2 = 1, which is a circle of radius 1. Substituting x for t in the y equation of II and squaring gives x2 + y2 = 1, but y ≥ 0 so this is only a semicircle. Squaring and substituting x2 for s in the y equation of III gives x2+ y2 = 1, but x ≥ 0 and so this is only a semicircle.

Piecewise Functions

* 1. (B) Enter abs(2x – 1) into Y1 and 4x + 5 into Y2. It is clear from the standard window that the two graphs intersect only at one point.

* 2. (E) Enter abs(x – 2) into Y1, 1 into Y2, and 4 into Y3. An inspection of the graphs shows that the values of x for which the graph of Y1 is between the other two graphs are in two intervals. E is the only answer choice having this configuration.

* 3. (A) Subtract |x| from both sides of the equation. Since |y| cannot be negative, graph the piecewise function

o15

  In the first command, the word “and” is in TEST/LOGIC. The result is a square that is o16 on a side. Therefore, the area is 8.

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4. (A) Since f(x) must = f(|x|), the graph must be symmetric about the y-axis. The only graph meeting this requirement is Choice A.

* 5. (B) Since the point where a major change takes place is at (1,1), the expression in the absolute value should equal zero when x = 1. This occurs only in Choice B. Check your answer by graphing the function in B on your graphing calculator.

6. (E) Choice A fails if N = 0.5. Choice B subtracts cents from ounces. Choice C fails if N = 1. Choice D adds cents to ounces.

7. (C) Since f(x) = an integer by definition, the answer is Choice C.

* 8. (E) Enter int(4x) – 2x into Y1. The graph is shown in the figure below.

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  The breaks in the graph indicate that it cannot be the graph of any of the first four answer choices.