SAT 8

134 PART II / MATH REVIEW

E X A M P L E :

Solve cos2 x = sin2 x over the interval 0° ≤ x ≤ 360°.

Assuming cos x ≠ 0, divide both sides of the equation by cos2 x.

1

= sin2 x

cos2 x 1 = tan2 x

tan x = 1 or tan x = −1

x = tan−1 (1) = 45°. On the interval 0° ≤ x ≤ 360°, 180 + 45 = 225° is also a solution.

x = tan−1 (−1) = −45°. On the interval 0° ≤ x ≤ 360°, 180 + −45 = 135° and 180 + 135 = 315° are solutions.

x = 45°, 135°, 225°, or 315° (Answer)

As mentioned, trigonometric equations can have infinitely many solutions. Since the period of the tangent function is 180° (or π radians), 45° plus any multiple of 180°, and 135° plus any multiple of 180° are possible solutions to the equation in the previous example.

E X A M P L E :

Solve 6sin2 x + sin x − 1 = 0 over the interval 0° ≤ x ≤ 360°.

Factor the equation to get:

(3sin x − 1)(2sin x + 1) = 0

sin x =	1		or sin x = −				1
	3						2
				1
x = sin−1						= 19.5°
				3
					1
x = sin−1			−			= −30°
					2

Considering 0° ≤ x ≤ 360°, the possible solutions are: x = 19.5°, 160.5°, 210°, or 330° (Answer)

*DOUBLE ANGLE FORMULAS

The double angle formulas for sine and cosine are as follows:

sin2θ = 2sinθcosθ cos2θ = cos2θ − sin2θ

Using the trigonometric identity sin2x + cos2x = 1, the double angle formula for cosine can also be written as:

cos 2θ = 1 − 2sin2 θ or cos 2θ = 2 cos2 θ − 1 The double angle formula for tangent is:

tan2θ = 2tanθ 1 − tan2θ

CHAPTER 7 / TRIGONOMETRY

135

E X A M P L E :

Given θ is in the first quadrant, if cos θ = 4, what is the value of sin 2θ? 5

If cos θ = 4, then sin θ = 3. (Sketch a 3-4-5 right triangle in quadrant I if

			5			5
		you’re unsure of the value of sin θ.)
	sin 2θ = 2 sin θ cos θ
			3			4
	sin 2θ = 2
			5			5
	=	24	(Answer)
		25

E X A M P L E :

Express 2 sin 3x cos 3x as a single trigonometric function.

Since the double angle formula for sine is sin 2θ = 2 sin θ cos θ,

2 sin 3x cos 3x = sin 2(3x)

= sin 6x (Answer)

E X A M P L E :

What is the value of 2 cos2 30° − 1?

Since 2 cos2 θ − 1 = cos 2θ,

2 cos2 30° − 1 = cos 2(30°) = cos 60°

cos 60° = 1	(Answer)
2

CHAPTER 8

FUNCTIONS

This chapter provides a review of functions. On the Level 2 test, 48–52% of the questions relate to Algebra and Functions combined. That translates to about 30% of the test questions relating specifically to Functions and about 20% to Algebra. The pie chart shows approximately how much of the Level 2 test is related to functions.

Numbers and

Operations

12% Algebra

20%

Data Analysis,

Statistics, and

Probability

8%

Solid

Geometry

4%

Coordinate

Geometry

12%

Functions

30%

Trigonometry

14%

The following topics are covered in this chapter:

1.Function Notation

2.Functions vs. Relations

a.Graphing Functions

3.Composition of Functions

a.Identity, Zero, and Constant Functions

4.Determining the Maximum or Minimum

5.The Roots of a Quadratic Function

6.Inverse Functions

7.Rational Functions

8.Higher-Degree Polynomial Functions

9.Exponential Functions

136

CHAPTER 8 / FUNCTIONS

137

10.*Logarithmic Functions

11.*Trigonometric Functions

12.*Inverse Trigonometric Functions

13.*Periodic Functions

14.*Piecewise Functions

15.*Recursive Functions

16.*Parametric Functions

FUNCTION NOTATION

A function is a set of ordered pairs of real numbers. A function is often described by a rule that creates a one-to-one correspondence between sets of input and output values. The set of all input values is called the domain of the function, and the set of all output values is called the range. Unless otherwise noted on the SAT Subject Test, the domain and range are assumed to be the set of all real numbers. Typically, a function is denoted by f (x) and is read “f of x” or “the value of the function f at x.” Functions can be represented by any letter, not just f, though. It is common to use f, F, g, G, h, and H to represent different functions. The following are examples of functions:

f (x) = x2. F(x) = 3x. g(x) = (x − 4)2 + 1.

Take the first example, f (x) = x2. When the function is evaluated for x = 2, the result is f (2) = 22 = 4. f (x) represents the y value when a function is graphed on a xy-coordinate plane. The point (2, 4) would be included in the graph of the function f (x) = x2 because when x = 2, f (x) = 4.

For a set of ordered pairs to be a function:

1.For each domain value of a function, there must be an associated range value.

2.A domain value cannot be matched with more than one range value.

When determining the domain of a function, look for two common restrictions: dividing by zero and identifying when the radicand is negative. You want to ensure that both cases do not happen.

A linear function is a function in the form: f (x) = mx + b where x, m, and b are real numbers.

The graph of a linear function is a straight line with slope m and y-intercept b.

A quadratic function is a function in the form: f (x) = ax2 + bx + c where a ≠ 0.

The graph of a quadratic function is a parabola.

*Denotes concepts that are on the Level 2 test only.

138	PART II / MATH REVIEW

A function is even if its graph is symmetric with respect to the y-axis, and odd if its graph is symmetric with respect to origin. To test for the symmetry of a function:

1.The function is even if replacing x with −x results in the original function. f (− x) = f (x).

2.The function is odd if replacing x with −x results in the opposite of the original function.

f (− x) = − f (x).

E X A M P L E :

Which of the following mapping diagrams depicts a function?

I.	D	R	II.	D	R
	0	0		0	5
	1	2		1	10
	2	4		2	15
	3	6		3	20
	4	8		4	25
III.	D	R	IV.	D	R
	0	–2		0
	1	–1		1	0
	2	0		2	2
	3	1		3	4
	4	2		4

(A)I and II

(B)II and III

(C)III and IV

(D)I and IV

(E)II and IV

Remember that there must be a range associated with each domain value. II is not a function because 4 doesn’t have an associated range. III cannot be a function because 2 is matched with both −1 and 0. It is ok for a range value to be mapped to more than one domain, such as shown in diagram IV. I and IV are functions. D is the correct answer.

CHAPTER 8 / FUNCTIONS	139
E X A M P L E :
Find the domain of f (x) =	100 − x2 .

The domain is the set of x values for which the function is defined. In other words, it is the set of values allowed for x. We are assuming that the domain of the function is the set of real numbers. (On the SAT Subject Test, you can assume this, unless told otherwise.) The radicand, 100 − x2, must, therefore, be positive.

100 − x2 ≥ 0.

100 ≥ x2.

10 ≥ x and 10 ≥ − x.

10 ≥ x and −10 ≤ x.

−10 ≤ x ≤ 10.

(Answer)

E X A M P L E :

Given f (x) = 2x − 5, find f (x) = x − 3.

Replace x in the original function with the expression x − 3.

f (x − 3) = 2(x − 3) − 5.
f (x − 3) = 2x − 6 − 5.
f (x − 3) = 2x − 11.		(Answer)
E X A M P L E :
Determine the domain of the function g(x) =				1	.
				x(x + 3)
The fraction	1		is not defined when the denominator x(x + 3)
	x(x + 3)

equals zero. Dividing by zero is a common restriction of the domain of a function.

x(x + 3) ≠ 0.
x ≠ 0 or x ≠ −3.
The domain is the set of all real numbers except 0 and −3.	(Answer)

140	PART II / MATH REVIEW

E X A M P L E :

Find the domain and range of the function f (x) = x − 1 shown in the graph.

y

x

It is useful to know that “v-shaped” graphs are associated with absolute value. The domain is the set of x for which the function is defined. Are there any restrictions on x in this example? The graph shows that x is defined at zero and at all positive and negative values. The domain is the set of all real numbers. What about the range? Looking at the equation, you see that y = x − 1 . Because y equals the absolute value of an expression, and absolute value, by definition, is always positive, y must be greater than or equal to zero. You can determine this from the graph because there are no points in either quadrant III or IV where y is negative.

D = {all real numbers}, R = { y: y ≥ 0}.

(Answer)

FUNCTIONS VS. RELATIONS

You can test for a function algebraically or by using the vertical line test. The vertical line test states that any vertical line intersects the graph of a function at, at most, one point. Both tests determine if there is one and only one range (y-value) associated with a given domain (x-value).

E X A M P L E :

Is y2 = 5 + x a function?

Let’s test this algebraically. Start by solving for y. y2 = 5 + x.

y = ± 5 + x.

You can immediately see that for one value of x there are two corresponding values of y. Let x = 4, for example. y = ± 5 + 4 , so y = ±3.

y2 = 5 + x is not a function. (Answer)

CHAPTER 8 / FUNCTIONS

141

E X A M P L E :

Is x2 + y2 = 4 a function?

Let’s test this one using the vertical line test. You may be able recognize that the graph is a circle because x2 + y2 = 4 fits into the standard form of the equation of a circle: (x − h)2 + (y − k)2 = r2. It center is (0,0), and its radius is 2. A vertical line intersects the circle at two points, as shown below.

y

x

The equation does not pass the vertical line test.

x2 + y2 = 4 is not a function.

(Answer)

By definition, a relation is any set of ordered pairs. A function, therefore, is simply a type of relation consisting of ordered pairs with different x coordinates. The two parabolas below show the difference between a relation and a function.

y

y

x

x

y = x2

x = y2

The graph of y = x2 passes the vertical line test, so it is a function. The graph of x = y2 does not pass the vertical line test (when x = 4, y = 2 or y = −2.), so it is a relation.

142	PART II / MATH REVIEW

E X A M P L E :

Which of the following is NOT a function?

(A){(x,y) : y = x}.

(B){(x,y) : x = y }.

(C){(x,y) : x2 = y}.

(D)	{(x,y) : y = x }.
(E)	{(x,y) : y = x3}.

Each answer is written as a set notation and represents a set of ordered pairs. Answer A, for example, is read “the set of all ordered pairs (x,y) such that y = x.”

(A)y = x is a linear equation and passes the vertical line test.

(B)x = y is a “v-shaped” graph that is concave left and does not pass the vertical line test. When x = 5 for example, y = ±5.

(C)x2 = y is a parabola whose vertex is the origin and is concave up, so it passes the vertical line test.

(D)y = x is a curve starting at the origin and moving upward in the positive x direction. Because it is a square root equation, it is not defined when x is negative. The range, y, is restricted to positive values because y equals a square root.

(E)y = x3 is a cubic curve that passes through the origin. The domain is the set of all real numbers, and each x has one y associated with it.

B is not a function.

(Answer)

Graphing Functions

The most straightforward way to graph a function is to create an xy-table and find the coordinates of points for which the function is true. Start with the intercepts; set y = 0 and solve for x, then set x = 0 and solve for y. Knowing the graphs of a few common functions may save you time on the SAT Subject Test. Be familiar with graphing the following functions:

•Identify function f (x) = x.

•Constant function f (x) = c.

•Absolute value function f (x) = x .

•Squaring function f (x) = x2.

• Square root function f (x) = x .

• Cubing function f (x) = x3.

CHAPTER 8 / FUNCTIONS

143

COMPOSITION OF FUNCTIONS

Two functions can be combined to form a composition. f (g(x)) is read as “the composition of f with g.” Let f (x) = x2 and g(x) = x + 1, evaluate the following:

f (g(x)) = f (x + 1) = (x + 1)2. g( f (x)) = g(x2 ) = x2 + 1.

f ( f (x)) = f (x2 ) = x4 . f (g(2)) = f (3) = 9.

Notice that f ( g(x)) is not equivalent to g( f (x)) in this example.

E X A M P L E :

Given g(x) = x + 3, h(x) = 9 − x2, and g(h(x)) = 8, find x.

g(h(x)) = g(9 − x2 ) = 9 − x2 + 3 = 12 − x2.

You’re given that g(h(x)) = 8, so set 12 − x2 equal to 8 and solve for x.

12 − x2 = 8.

4 = x2.
x = ± 2.	(Answer)

E X A M P L E :

2

Given f (x) = x3 and g(x) = x6. Find f (g(8x)). Start by finding g(x).

g(8x) = (8x)6. Now find f ((8x)6).

	2
f ((8x)6 ) = [(8x)6 ]			= 84 x4.
		3
= 4096x4.	(Answer)

Identity, Zero, and Constant Functions

The identity function is the function for which y = x. f (x) = x. Its graph is a diagonal line passing through the origin whose slope is 1.

The zero function is the function that assigns 0 to every x. f (x) = 0. Its graph is the horizontal line in which y = 0, otherwise known as the x-axis.

A constant function is any function that assigns a constant value c to every x. f (x) = c. Its graph is a horizontal line, y = c, whose y-intercept is the point (0,c).