SAT 8

are the same.

154 PART II / MATH REVIEW

E X A M P L E :

Determine if x = 2 is a zero of the function g(x) = x3 − 7x + 6.

If x = 2 is a zero of the function, then x − 2 must be a factor of the polynomial. By the factor theorem, if g(2) = 0, then x − 2 is, in fact, a factor of g.

g(2) = (2)3 − 7(2) + 6 = 8 − 14 + 6 = 0.

2 is a zero of g(x). (Answer)

EXPONENTIAL FUNCTIONS

An exponential function f with base a is given by: f (x) = ax

where x is a real number, a > 0 and a ≠ 1.

The graphs of y = ax and y = a−x are as follows:

												y																																						y
										8																																						8
										6																																						6
										4																																						4
										2																									x													2																																x
			–5																						5														–5																												5
										–2																																						–2
											y = ax																																				y = a–x

Notice that both have a y-intercept of 1 and a horizontal asymptote at y = 0. The graph of y = a−x is a reflection of the graph of y = ax over the y-axis.

E X A M P L E :

Use the properties of exponents to determine if the functions f (x) = 27(3−x)

and g(x) = 1 x− 3

3

Review the properties of exponents given in the Algebra chapter if you’re

unsure of how to solve this problem.
f (x) = 27(3− x ) = 33 (3− x ) = 33− x.
1		x− 3
g(x) =			= (3−1 )x− 3 = 3− x+ 3
	3
f (x) and g(x) are the same functions.				(Answer)

4. log4 4 2

5. log2 2

CHAPTER 8 / FUNCTIONS

155

E X A M P L E :

If f (x) = 2x and g(x) = 3x, then determine when 3x < 2x.

Both functions have a y-intercept of 0. Graph the functions on your graphing calculator (use the ^ button to raise 2 and 3 to the x power) to compare the curves. When x > 0, 3x > 2x.

3x < 2x when x < 0. (Answer)

*LOGARITHMIC FUNCTIONS

A logarithmic function with base a is given by: f (x) = loga x

where x > 0, a > 0, and a ≠ 1.

If y = loga x then x = ay. A logarithm is an exponent. For example, log2 16 is the exponent to which 2 must be raised to get 16. Therefore, log2 16 = 4.

Try evaluating the following logarithms:

1.log3 81 = 4

2.log10 0.1 = −1

3.log2 (−4) = undefined

−5 = − 52 2 = 32

6.log9 3 = 12

7.log5 1 = 0.

The properties of logarithms are:

1.loga 1 = 0

2.loga a = 1

3.loga ax = x

4.loga ( pq) = log a p + log a q

5.loga p = log a p − log a q

q

6.loga px = x log a p

7.alog a p = p

Logarithmic functions in base 10 are called common logarithmic func-

tions. The log button on your calculator calculates common logs. For example, try entering log 100 into your calculator. Although the base is not written, it is assumed to be base 10. log10 100 = 2.

156	PART II / MATH REVIEW

The Change of Base Formula states that:

loga x = logb x , logb a

where a, b, and x are positive real numbers and a and b ≠ 1.

With the Change of Base Formula, you can use your calculator to evaluate logs in any base, not just base 10.

log2 7 = log10 7 = 2.807. log10 2

log3 35 = log10 35 = 3.236. log10 3

E X A M P L E :

Simplify log5 150 − log5 6.

log5 150 − log5 6 = log5 1506 = log5 25 = 2.

2.(Answer)

E X A M P L E :

Simplify log4 2 + log4 8.

log4 2 − log4 8 = log4 2(8) = log4 16 = 2.

2.(Answer)

E X A M P L E :

											N6				1
															3
Use the properties of logarithms to express log b																in terms of log b M
												M	4
and log b N.
	N6		1				1		N6
logb =				3		=
								logb			.
	M4						3
									M4

=13 (logb N6 − logb M4 ).

=13 (logb N6 − logb M4 ) = 13 (6logb N − 4logb M).

= 2logb N −	4 logb M.	(Answer)
	3

E X A M P L E :

Use the properties of logarithms to express log (x + 2) + 2log x − log (x − 4) as a single logarithm.

log(x + 2) + 2log x − log(x − 4) = log(x + 2) + log x2 − log(x − 4)

= log	(x + 2)x2 .	(Answer)
	x − 4

CHAPTER 8 / FUNCTIONS

157

E X A M P L E :

Solve log (x − 1) − log (x + 1) = log (x + 5). log (x − 1) − log (x + 1) = log (x + 5).

x − 1

log x + 1 = log (x + 5). This is true only when:

x − 1 = x + 5. x + 1

x − 1 = (x + 5)(x + 1). x − 1 = x2 + 6x + 5.

0= x2 + 5x + 6.

0= (x + 3)(x + 2). x = −3 or x = −2.

But the logarithm function is undefined for both x = −3 and x = −2.

No solution. (Answer)

The logarithmic function is the inverse of the exponential function.

Recall that inverse functions are reflections of each other over the line y = x. Because of the nature of inverses, exponential equations can be solved by isolating the exponential expression and taking the logarithm of both sides. Logarithmic equations can be solved by rewriting the equation in exponential form and solving for the variable.

E X A M P L E :

						x			= 11.
	Solve 6 log3
							2
		x			= 11.
	6 log3
			2
	x			=	11
	log3						.
		2			6

Write the equation in exponential form to get:

11
3 6 = x .
2
2(3116 ) = x.
x = 14.99.	(Answer)

158 PART II / MATH REVIEW

E X A M P L E :

Solve 64x = 1,000.

Take the log of both sides to get: log64 x = log 1,000.

	4xlog 6 = 3.
	3
	x =
		4log 6
	x = 0.964.							(Answer)
	E X A M P L E :
	Solve 3x−1			=		1			.
						729
	Note that 729 can be written in base 3.
	1					1		6
	3x−1 =			=					= −3−6.
			729			3

3x−1 = −3−6.

When the bases are the same, simply set the exponents equal and solve for x. (You could also solve this equation by taking the log of both sides.)

x− 1 = −6.

x= −5. (Answer)

CHAPTER 8 / FUNCTIONS

159

*TRIGONOMETRIC FUNCTIONS

Trigonometric ratios, cofunctions, identities, and equations are covered in the Trigonometry chapter. Here we discuss trigonometric graphs. The graphs of all six trigonometric functions (sine, cosine, tangent, cosecant, secant, and cotangent) are periodic, meaning that the values of the functions repeat themselves at regular intervals. The sine, cosine, cosecant, and secant functions have periods of 2π, while the tangent and cotangent functions have periods of π. The graphs of sine and cosine are as follows:

y

2

f(x) = sin(x)

x

–5

5

–2

y

2

f(x) = cos(x)

x

–5

5

–2

Sine is an odd function because it is symmetric with respect to the origin. Cosine is an even function because it is symmetric with respect to the y-axis. Notice that both curves are the same shape with the cosine curve equivalent to

π

the sine curve shifted 2 units left. Both curves also have a domain of all real numbers and a range of −1 ≤ y ≤ 1.

160	PART II / MATH REVIEW

In general, the sine and cosine functions can be written as: y = d + a sin (bx − c) and y = d + a cos (bx − c),

where d represents a vertical shift in the graph, a is the amplitude (the vertical stretching or shrinking), c is a horizontal shift in the graph, and b is the hor-

izontal stretching or shrinking. 2bπ is the period of the function. The amplitude

can also be thought of as half of the vertical distance between the maximum and minimum points on the graph, while the vertical shift can be thought of as the average of the maximum and minimum values. Use c to solve for the right and left endpoints of one cycle of the graph by solving the equations bx − c = 0 and bx − c = 2π.

E X A M P L E :

Determine the amplitude and period of y = 3 cos

The amplitude is 3 and the period is 2π = 4. π /2

π

2 x.

(Answer)

E X A M P L E :

For what 2π = 2π b k k = 82ππ =

value of k is the period of the function y = 12 sin kx equal to 8π?

= 8π.

14 . (Answer)

E X A M P L E :

Find an equation in the form y = d + a sin (bx − c) that has a maximum of 9 and minimum of 1 and a period of π.

The amplitude equals half of the vertical distance between the maximum and minimum points.

a = (9 − 1) = 4. 2

The vertical shift equals the average of the maximum and minimum values.

d =			(9 + 1)	= 5.
		2
2π		= π, so b = 2.
b
y = 5 + 4 sin 2x.					(Answer)

(Note that y = 5 + 4 sin 2x is one possible equation where there is no horizontal shift in the graph and a > 0. There are, however, other possible answers.)

to the origin. It has a period of π and has asymptotes. Because tan x =

CHAPTER 8 / FUNCTIONS

161

The tangent function is an odd function because it is symmetric with respect

sin x cos x ,

it is undefined when cos x = 0. This occurs when x = π2 or any odd multiple of

π2 . As x approaches π2 , the tangent of x increases without bound. The graph of the tangent function is as follows:

y

4

f(x) = tan(x)

2

x

–5

5

–2

The domain of the tangent function is all real numbers except odd multiples of π2 , and its range is the set of all real numbers.

As with the graphs of sine and cosine, the tangent function can be written

as:

y = d + a tan (bx − c),

where d represents a vertical shift in the graph, a is the vertical stretching or shrinking, c is a horizontal shift in the graph, and b is the horizontal stretching

or shrinking. πb is the period of the function. Note that amplitude of a tangent function is undefined.

162	PART II / MATH REVIEW

The graphs of the reciprocal functions, cosecant, secant, and cotangent, are shown below. The cosecant and cotangent functions have asymptotes at all multiples of π, while the secant function has asymptotes at all odd multiples

of π2 .

y

y = csc x

2

x

–5

5

–2

y

2

y = sec x

x

–5

5

–2

y

y = cot x

2

x

–5

5

–2

	CHAPTER 8 / FUNCTIONS									163
	E X A M P L E :
		What is the period of y = sec								1	x + 1?
										2
		2π	=	2π				= 4π.	(Answer)
		b			1
			2

E X A M P L E :

For which values over the period 0 ≤ x < 2π is the function y = 3cot x undefined?

The cotangent is defined for all real numbers except multiples of π. It is undefined at 0 and π. (Answer)

*INVERSE TRIGONOMETRIC FUNCTIONS

The inverse trigonometric functions, arcsin (sin−1), arccos (cos−1), and arctan (tan−1), represent angle measures. As with any inverse function, their graphs are obtained by reflecting the original trigonometric functions over the line y = x. Recall that a function has an inverse if it passes the Horizontal Line Test. The trigonometric functions obviously do not pass this test, but they do have inverses on restricted domains.

The domain of y = sin−1 x is −1 ≤ x ≤ 1, and the range is − π2 ≤ y ≤ π2 .

The domain of y = cos−1 x is −1 ≤ x ≤ 1, and the range is 0 ≤ y ≤ π.

The domain of y = tan−1 x is all real numbers, and the range is − π2 ≤ y ≤ π2 .

The graphs of these three inverse trigonometric functions are as follows (Figure 8-12):