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

Part 2. REVIEW OF MAJOR TOPICS

This part reviews the mathematical concepts and techniques for the topics covered in the Math Level 2 Subject Test. A sound understanding of these concepts certainly will improve your score. The techniques discussed may help you save time solving some of the problems without a calculator at all. For problems requiring computational power, techniques are described that will help you use your calculator in the most efficient manner.

Your classroom experience will guide your decisions about how best to use a graphing calculator. If you have been through a secondary mathematics program that attached equal importance to graphical, tabular, and algebraic presentations, then you probably will rely on your graphing calculator as your primary tool to help you find solutions. However, if you went through a more traditional mathematics program, where algebra and algebraic techniques were stressed, it may be more natural for you to use a graphing calculator only after considering other approaches.

Chapter 1. Functions

1.1 Overview

DEFINITIONS

A function is a process that changes a set of input numbers into a set of output numbers. Functions are usually specified by equations such as . In this equation x represents an input number while y represents the (unique) corresponding output number. Functions can also have names: in the example, we could name the function f. Then the process could be described as , whereby f takes the input number x, multiplies it by 2, subtracts 1, and takes the square root to produce the output y = f (x).

Taken as a group, the input numbers are called the domain of the function, while the output numbers are called the range. Unless otherwise specified, the domain of a function is all real numbers for which the equation produces outputs that are real numbers. In the example above, the domain is the set x ≥ . since 2x − 1 cannot be negative if y is to be a real number. In this case, the range is the set of all non-negative numbers.

The domain of a function can also be established as part of the definition of a function. For example, even though the domain of the example function f is x ≥ , one could, for example, specify the domain x > 5. Unless a domain is explicitly stated, the domain is assumed to be all real values that produce real numbers as outputs.

A function with a small finite domain can be described by a set of ordered pairs instead of an equation. The first number in the pair is from the domain and the second is the corresponding range value. Consider, for example, the function f consisting of the pairs (0, 2), (1, 1), (2, 3), and (3, 8). In this example, the “process” is not systematic: it simply changes 0 to 2 (f (0) = 2); 1 to 1 (f (1) = 1); 2 to 3 (f (2) = 3); and 3 to 8 (f (3) = 8). The domain of this function consists of 0, 1, 2, and 3, while the range consists of 2, 1, 3, and 8. Functions like this are typically used to illustrate certain properties of functions and are discussed later.

A function is actually a special type of relation. A relation describes the association between two variables. An equation such as x² + y² = 4 is one way of defining a relation. All ordered pairs (x, y) that satisfy the equations are in the relation. In this case, these pairs form the circle of radius 2 centered at the origin.

Circles are examples of relations that are not functions because some x values (0 in the example) have two y values associated with it) (2 and −2), which violates the uniqueness of the output for a given input. Other than circles, relations that are not functions include ellipses, hyperbolas, and parabolas that open right or left, instead of up or down—in other words, the conic sections discussed in Section 2.3.

Like functions, relations can also be defined using specific ordered pairs. The set R = {(1, 1), (1, 2), (2, 1), (2, 2)} is an example of a relation that is not a function because the x value 1 has two y values associated with it (1 and 2).

EXERCISES

1. If {(3,2),(4,2),(3,1),(7,1),(2,3)} is to be a function, which one of the following must be removed from the set?

  (A) (3,2)

  (B) (4,2)

  (C) (2,3)

  (D) (7,1)

  (E) none of the above

2. For f(x) = 3x² + 4, g(x) = 2, and h = {(1,1), (2,1), (3,2)},

  (A) f is the only function

  (B) h is the only function

  (C) f and g are the only functions

  (D) g and h are the only functions

  (E) f, g, and h are all functions

3. What value(s) must be excluded from the domain of ?

  (A) –2

  (B) 0

  (C) 2

  (D) 2 and –2

  (E) no value

COMBINING FUNCTIONS

Given two functions, f and g, five new functions can be defined:

EXAMPLE

If f(x) = 3x – 2 and g(x) = x² – 4, write an expression for each of the following functions:

 (A) (f + g)(x)

 (B) (f – g)(x)

 (C) f · g (x)

 (D)

 (E) (f g)(x)

 (F) (g f )(x)

SOLUTIONS

  (A)

  (B)

  (C)

  (D)

  (E)

  (F)

EXERCISES

1. If f(x) = 3x² – 2x + 4, f(–2) =

  (A) –12

  (B) –4

  (C) –2

  (D) 12

  (E) 20

2. If f(x) = 4x – 5 and g(x) = 3^x, then f(g(2)) =

  (A) 3

  (B) 9

  (C) 27

  (D) 31

  (E) none of the above

3. If f(g(x)) = 4x² – 8x and f(x) = x² – 4, then g(x) =

  (A) 4 – x

  (B) x

  (C) 2x – 2

  (D) 4x

  (E) x²

4. What values must be excluded from the domain of (x) if f(x) = 3x² – 4x + 1 and g(x) = 3x² – 3?

  (A) 0

  (B) 1

  (C) 3

  (D) both ±1

  (E) no values

5. If g(x) = 3x + 2 and g(f(x)) = x, then f(2) =

  (A) 0

  (B) 1

  (C) 2

  (D) 6

  (E) 8

6. If p(x) = 4x – 6 and p(a) = 0, then a =

  (A) −6

  (B)

  (C)

  (D)

  (E) 2

7. If f(x) = e^x and g(x) = sin x, then the value of (f g)( ) is

  (A) –0.01

  (B) –0.8

  (C) 0.34

  (D) 1.8

  (E) 2.7

INVERSES

The inverse of a function f, denoted by f ^–1, is a relation that has the property that f(x) f ^–1(x) = f ^–1(x) f(x) = x. The inverse of a function is not necessarily a function.

EXAMPLES

1. the inverse of f ?

To answer this question assume that and verify that f(x) f ^–1(x) = x.

To verify this, proceed as follows:

and

Since is the inverse of f.

2. f = {(1,2),(2,3),(3,2)}. Find the inverse.

   f ⁻¹ ={(2,1), (3,2), (2,3)}

To verify this, check f f ^–1 and f ^–1 f term by term.

Thus, for each x, f(f ^–1(x)) = x.

Thus, for each x, f ^–1(f(x)) = x. In this case f ^–1 is not a function.

If the point with coordinates (a,b) belongs to a function f, then the point with coordinates (b,a) belongs to the inverse of f. Because this is true of a function and its inverse, the graph of the inverse is the reflection of the graph of fabout the line y = x.

3. f ^–1 is not a function.

4. f ^–1 is a function.

As can be seen from the above examples, the graph of an inverse is the reflection of the graph of a function (or relation) through the line y = x. Algebraically, the equation of an inverse of a function can be found by replacing f (x) by y; interchanging x and y; and solving the resulting equation for y.

5. f(x) = 3x + 2. Find f ⁻¹.

In order to find f ^–1, interchange x and y and solve for y: x = 3y + 2, which becomes .

Thus,

6. f(x) = x². Find f ⁻¹.

Write y = x²

Interchange x and y: x = y².

Solve for

Thus, the inverse of y = x² is not a function.

The inverse of any function f can always be made a function by limiting the domain of f. In Example 6 the domain of f could be limited to all nonnegative numbers or all nonpositive numbers. In this way f ^–1 would become either or , both of which are functions.

7. f(x) = x² and x ≥ 0. Find f ⁻¹.

Write y = x² and x ≥ 0. Then switch x and y: x = y² and y ≥ 0.

Solve for y: .

Here f ⁻¹ is the function

Finding an equation for the inverse of a function can also be used to determine the range of a function, as shown in the following example.

8. Find the range of .

First replace f(x) by y, and interchange x and y to get . Then solve for y:

In order for this to be defined, x ≠ −2. In other words, −2 is not in the range of f. (This could also be determined by observing that in the original function can never be zero.)

EXERCISES

1. If f(x) = 2x – 3, the inverse of f, f ^–1, could be represented by

  (A) f ^–1(x) = 3x − 2

  (B)

  (C)

  (D)

  (E)

2. If f(x) = x , the inverse of f, f ^–1, could be represented by

  (A)

  (B)

  (C)

  (D)

  (E) f ^–1 does not exist

3. The inverse of f = {(1,2),(2,3),(3,4),(4,1),(5,2)} would be a function if the domain of f is limited to

  (A) {1,3,5}

  (B) {1,2,3,4}

  (C) {1,5}

  (D) {1,2,4,5}

  (E) {1,2,3,4,5}

4. Which of the following could represent the equation of the inverse of the graph in the figure?

  (A) y = –2x + 1

  (B) y = 2x + 1

  (C)

  (D)

  (E)

ODD AND EVEN FUNCTIONS

A relation is said to be even if (–x,y) is in the relation whenever (x,y) is. If the relation is defined by an equation, it is even if (–x,y) satisfies the equation whenever (x,y) does. If the relation is a function f, it is even if f(–x) = f(x) for all x in the domain of f. The graph of an even relation or function is symmetric with respect to the y axis.

EXAMPLES

1. {(1,0),(–1,0),(3,0),(–3,0),(5,4),(–5,4)} is an even relation because (–x,y) is in the relation whenever (x,y) is.

2. x⁴ + y² = 10 is an even relation because (–x)⁴ + y² = x⁴ + y² = 10.

3. f(x) = x² and f(–x) = (–x)² = x².

4. f(x) = | x | and f(−x) = | −x | = | −1 · x | = | −1 | · | x | = | x |.

A relation is said to be odd if (–x,–y) is in the relation whenever (x,y) is. If the relation is defined by an equation, it is odd if (–x,–y) satisfies the equation whenever (x,y) does. If the relation is a function f, it is odd if f(–x) = –f(x) for all x in the domain of x. The graph of an odd relation or function is symmetric with respect to the origin.

5. {(5,3),(–5,–3),(2,1),(–2,–1),(–10,8), (10,–8)} is an odd relation because (–x,–y) is in the relation whenever (x,y) is.

6. x⁴ + y² = 10 is an odd relation because (–x)⁴ + (–y)² = x⁴ + y² = 10. Note that x⁴ + y² = 10 is both even and odd.

7. f(x) = x³ and f(–x) = (–x)³ = –x³.

Therefore, f(–x) = x³ = –f(x).

8. and .

Therefore, .

The sum of even functions is even. The sum of odd functions is odd. The product of an even function and an odd function is odd. The product of two even functions or two odd functions is even.

EXERCISES

1. Which of the following relations are even?

  I. y = 2

  II. f(x) = x

  III. x² + y² = 1

  (A) only I

  (B) only I and II

  (C) only II and III

  (D) only I and III

  (E) I, II, and III

2. Which of the following relations are odd?

  I. y = 2

  II. y = x

  III. x² + y² = 1

  (A) only II

  (B) only I and II

  (C) only I and III

  (D) only II and III

  (E) I, II, and III

3. Which of the following relations are both odd and even?

  I. x² + y² = 1

  II. x² – y² = 0

  III. x + y = 0

  (A) only III

  (B) only I and II

  (C) only I and III

  (D) only II and III

  (E) I, II, and III

4. Which of the following functions is neither odd nor even?

  (A) {(1,2),(4,7),(–1,2),(0,4),(–4,7)}

  (B) {(1,2),(4,7),(–1,–2),(0,0),(–4,–7)}

  (C) y = x³ – 1

  (D) y = x² – 1

  (E) f(x) = –x

Answers and Explanations

Definitions

1. (A) Either (3,2) or (3,1), which is not an answer choice, must be removed so that 3 will be paired with only one number.

2. (E) For each value of x there is only one value
for y in each case. Therefore, f, g, and h are all functions.

3. (C) Since division by zero is forbidden, x cannot equal 2.

Combining Functions

1. (E) f(–2) = 3(–2)² – 2(–2) + 4 = 20.

2. (D) g(2) = 3² = 9. f(g(2)) = f(9) = 31.

3. (C) To get from f(x) to f(g(x)), x² must become 4x². Therefore, the answer must contain 2x since (2x)² = 4x².

4. (D) g(x) cannot equal 0. Therefore, .

5. (A) Since f(2) implies that x = 2, g(f(2)) = 2. Therefore, g(f(2)) = 3(f(2)) + 2 = 2. Therefore, f(2) = 0.

6. (C) p(a) = 0 implies 4a – 6 = 0, so .

* 7. (E)

Inverses

1. (E) If y = 2x – 3, the inverse is x = 2y – 3, which is equivalent to .

2. (A) By definition.

3. (B) The inverse is {(2,1),(3,2),(4,3),(1,4),(2,5)}, which is not a function because of (2,1) and (2,5). Therefore, the domain of the original function must lose either 1 or 5.

4. (E) If this line were reflected about the line y = x to get its inverse, the slope would be less than 1 and the y-intercept would be less than zero. The only possibilities are Choices D and E. Choice D can be excluded because since the x-intercept of f(x) is greater than –1, the y-intercept of its inverse must be greater than –1.

Odd and Even Functions

1. (D) Use the appropriate test for determining whether a relation is even.

  I. The graph of y = 2 is a horizontal line, which is symmetric about the y-axis, so y = 2 is even.

  II. Since f(–x) = –x x = f(x) unless x = 0, this function is not even.

  III. Since (–x)² + y² = 1 whenever x² + y² = 1, this relation is even.

2. (D) Use the appropriate test for determining whether a relation is odd.

  I. The graph of y = 2 is a horizontal line, which is not symmetric about the origin, so y = 2 is not odd.

  II. Since f(–x) = –x = –f(x), this function is odd.

  III. Since (–x)² + (–y)² = 1 whenever x² + y² = 1, this relation is odd.

3. (B) The analysis of relation III in the above examples indicates that I and II are both even and odd. Since –x + y 0 when x + y = 0 unless x = 0, III is not even, and is therefore not both even and odd.

4. (C) A is even, B is odd, D is even, and E is odd. C is not even because (–x)³ – 1 = –x³ – 1, which is neither x³ – 1 nor –x³ + 1.