SAT 8

164	PART II / MATH REVIEW

y

2

f(x) = sin–1 (x)

x

–2

y

2

g(x) = cos–1 (x)

x

–2

y

2

h(x) = tan–1 (x)

x

–2

As with any inverse function, the compositions f ( f−1(x)) = x and f−1( f (x)) hold true. For their respective domains and ranges:

sin(arcsin x) = x cos(arccos x) = x tan(arctan x) = x

sin(arcsin y) = y cos(arccos y) = y tan(arctan y) = y

	CHAPTER 8 / FUNCTIONS																					165
	E X A M P L E :
	Evaluate arcsin							2	.
								2
	The arcsin		2			represents the angle whose sine is											2	.
			2														2
	Because sin			π		=		2		, arcsin		2	=	π	.	(Answer)
			4					2				2
													4
	E X A M P L E :
	Evaluate arcsin(sin π).																			π		π
	arcsin(sin y) = y is only valid for values of y in the interval																		−		≤ y ≤		.
																						2
																			2
	π does not lie within the range of the arcsine function, but sin π = 0.
	arcsin(sin π) = arcsin(0) = 0.
	arcsin(sin π) = 0.									(Answer)
	E X A M P L E :
	Evaluate tan(arctan 4).
	The domain of tan(arctan x) = x is the set of all real numbers, so:
	tan(arctan 4) = 4.									(Answer)

E X A M P L E :

Evaluate sin(arcos 45 ).

This is asking for the sine of the angle whose cosine is 45 . Think about a

right triangle in quadrant I whose hypotenuse measures 5 and whose leg adjacent to the given angle measures 4.

The sine of the angle is	3.	(Answer)
	5

*PERIODIC FUNCTIONS

A function is periodic if for every x in its domain there is some constant c such that:

f (x + c) = f (x).

The smallest value of c is the period of the function. The sine and cosine functions, for example, are periodic and have a period of 2π.

166	PART II / MATH REVIEW

E X A M P L E :

Part of the graph of a function f having period 2 is shown below. Sketch the graph of f on the interval −4 ≤ x ≤ 4.

y

2

x

–2

The given part of the graph represents one period. Repeat the graph to the right and left until you reach 4 and −4, respectively. The resulting graph is:

y

2

x

–2

E X A M P L E :

Part of the graph of a function f having period 2 is shown below. Sketch the graph of f on the interval −2 ≤ x ≤ 2 given that the function is odd.

y

2

x

–2

+ 3 if x < 0.

CHAPTER 8 / FUNCTIONS

167

Recall that an odd function is symmetric with respect to the origin (versus an even function which is symmetric with respect to the y-axis.) Reflecting the given part of the function over the origin results in the following graph:

y

2

x

–2

*PIECEWISE FUNCTIONS

A piecewise function has different rules for different intervals of its domain.

E X A M P L E :

Graph the piecewise function:

12x

f (x) = − x + 3 if 0 ≤ x ≤ 3.3x − 9 if x > 3.

The piecewise function has three parts and all of them are linear equations. The graph looks like the following:

	y
	4
	2
	x
–5	5
	–2
	–4

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The greatest integer function, f (x) = [x], is another example of a piecewise function. f (x) equals the greatest integer less than or equal to x. f (1.5) = [1.5] = 1. f (−0.5) = [−0.5] = −1. The graph of the f (x) = [x] jumps up one unit at each integer and is a horizontal segment between consecutive integers. Because of its vertical breaks, the greatest integer function is also called a step function. Its graph is as follows:

y

2

x

–2

*RECURSIVE FUNCTIONS

A recursive function defines the terms in a sequence by relating each term to the previous ones.

The Fibonacci Sequence 1, 1, 2, 3, 5, 8, 13, . . . can be defined recursively as:

an = an− 2 + an−1 where n ≥ 2, a0 = 1 and a1 = 1.

A factorial can also be thought of as being recursive because n! = 1 × 2 × 3 × 4 . . . (n − 1) × n. In other words, n! = (n − 1)! × n.

E X A M P L E :

If fn = 2fn−1 + 3 for n = 2, 3, 4, . . . and f1 = 3, then what is f4?

Start by looking at the pattern formed by consecutive terms.

f1	= 3
f2	= 2f1 + 3 = 2(3) + 3 = 9.
f3	= 2f2	+ 3 = 2(9) + 3 = 21.
f4	= 2f3	+ 3 = 2(21) + 3 = 45.
f4	= 45.	(Answer)

CHAPTER 8 / FUNCTIONS

169

E X A M P L E :

If fn + 1 = fn−1 + 3fn for n = 2, 3, 4, . . . and f1 = 1 and f2 = 2, then what is f5?

Because you are given the first two terms of the sequence, you can define the other terms using these.

f1	= 1.
f2	= 2.
f3	= f1 + 3f2	= 1 + 3(2) = 7.
f4	= f2 + 3f3	= 2 + 3(7) = 23.
f5	= f3 + 3f4	= 7 + 3(23) = 76.
f5	= 76.	(Answer)

*PARAMETRIC FUNCTIONS

The content in this chapter has focused on graphs involving two variables, x and y, but it is also possible to express x and y in terms of a third variable. This third variable is called a parameter. Examples of parametric equations are as follows:

x = 12t.

y = −2t2 + 12t.

Substituting values for the parameter results in ordered pairs (x, y), which are points on the graph of the function. A curve represented by a pair of parametric equations is still plotted in an xy-coordinate plane. Plotting points as t increases gives the curve in a certain direction, which is called its orientation.

E X A M P L E :

Graph the curve determined by the parametric equations x = t2 − 1 and y = 3t on the interval 0 ≤ t ≤ 4.

Substitute values of t to determine points on the curve.

t	0	1	2	3	4
x	−1	0	3	8	15
y	0	3	6	9	12

170	PART II / MATH REVIEW

Now, plot the points in order to get the following graph:

y

8

6

4

2

x

–5

5

–2

Parametric equations can also be graphed by finding an equation in terms of x and y that has the same graph. When eliminating the parameter, check the domain of the new function, since the resulting graph may contain points that are not on the graph of the original parametric equations.

E X A M P L E :

Eliminate the parameter and write the rectangular equation whose graph represents the parametric curve x = t2 and y = 4t2 − 1.

Substitute x = t2 into the equation for y to get:

y = 4x − 1.

The equation is linear. The domains of the original parametric equations show that x ≥ 0 and y ≥ −1, however, because squaring t always results in a positive value. The domain of the rectangular equation must be x ≥ 0.

y = 4x − 1 and x ≥ 0.

(Answer)

CHAPTER 9

DATA ANALYSIS, STATISTICS, AND PROBABILITY

This chapter provides a review of elementary statistics. On the Level 2 test, 6–10% of the questions relate specifically to data analysis, statistics, and probability. The statistics problems on the Level 2 test focus on measures of central tendency, range, graphs and plots, regression, and basic probability. Counting problems are included in the Numbers and Operations chapter. The pie chart shows approximately how much of the Level 2 test is related to data analysis, statistics, and probability:

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.Mean, Median, Mode

2.Range

3.Interquartile Range

4.*Standard Deviation

5.Data Interpretation

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

171

172	PART II / MATH REVIEW

6.Regression

a.Linear Regression

b.*Quadratic Regression

c.*Exponential Regression

7.Probability

MEAN, MEDIAN, MODE

Three different statistics are commonly used to measure central tendency. They are:

Mean—the average of the numbers

Median—the middle number (when the data is ordered)

Mode—the number that occurs the most

The mean is calculated by finding the sum of all the terms and dividing by the total number of terms. After the data are ordered, the median is simply the middle value of an odd number of terms or the average of the two middle values for an even number of terms. The mode is the most frequent value. It is possible for data to have more than one mode.

E X A M P L E :

The ages of the starting players on a high school soccer team are as follows:

14, 15, 15, 16, 16, 16, 17, 17, 17, 17, 18

Find the mean, median, and mode of the data.

1.Mean—Calculate the mean by finding the sum of all the ages and dividing that by the number of ages in question.

The mean is

14 + 15 + 15 + 16 + 16 + 16 + 17 + 17 + 17 + 17 + 18 11

= 17811 ≈ 16.18.

2.Median—The 6th term of 11 total terms is the middle number. The median is 16.

3.Mode—17 occurs 4 times in the given data. The mode is 17.

Mean ≈ 16.18, Median = 16, Mode = 17.

(Answer)

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

CHAPTER 9 / DATA ANALYSIS, STATISTICS, AND PROBABILITY	173
E X A M P L E :

Sarah has test scores of 65, 78, 81, 82, and 90. What must she score on her 6th test to maintain an average score of 80?

Let x be Sarah’s 6th test score. Because you know the average, or mean, of the data, set up an equation equal to 80 and solve for x.

80 = 65 + 78 + 81 + 82 + 90 + x . 6

80 = 396 + x .

6

480 = 396 + x.

84 = x.

Sarah must score 84 on her 6th test.

(Answer)

E X A M P L E :

Find the median and mode of the following distribution:

0, 0, 1, 4, 6, 8, 8, 9

The mode is the number that occurs the most. Because both 0 and 8 occur twice, each number is a mode.

The median is the middle number. Because there’s an even number of terms in the given distribution, the median cannot be one of the given terms. Instead, find the median by adding the 4th and 5th terms and dividing by two. (In essence, you’re finding the average of the two terms because the middle occurs between them.) The median is:

	4 +	6	= 5.
2
The mode is 0 and 8, and the median is 5.				(Answer)

RANGE

Range is a simple measure of data dispersion. Range is the difference between the largest and smallest data values. Take Sarah’s test scores of 65, 78, 81, 82, and 90 from one of the previous examples. The range of her scores is 25. 90 − 65 = 25.

E X A M P L E :

Given the daily high temperatures for the week are: 61°, 65°, 76°, 69°, 60°, 72°, and 69°, what is the range of the data?

The highest temperature is 76° and the lowest temperature is 60°, so the range is:

76 − 60 = 16.

(Answer)