Cracking the SAT subject test in math 2, 2nd edition

Cracking the SAT Subject Test in Math 2, 2nd Edition CHAPTER 10 FUNCTIONS DRILL EXPLANATIONS Drill 1 Functions 19 B Just follow instructions on this one, and you get −64 − (−27), or −64 + 27, which is[.]

EXPLANATIONS

Drill 1: Functions

19 B Just follow instructions on this one, and you get −64 − (−27), or

−64 + 27, which is −37 (C) is a trap answer

20 A You’ve just got to plow through this one The original

expression ¥5 + ¥6 becomes 5(3)2 + 5(4)2, which equals 125 Work through the answer choices from the top to find the one that gives you 125 Choice (E) is a trap answer

21 B The function §a leaves even numbers alone and flips the signs

of odd numbers That means that the series §1 + §2 + §3…§100 + §101 will become (−1) + 2 + (−3) + 4 + (−5)…+ 100 + (−101) Rather than adding up all those numbers, find the pattern: −1 and 2 add up to 1; −3 and 4 add up to 1; and so on, all the way

up to −99 and 100 That means 50 pairs that add up to 1, plus the −101 left over 50 + −101 = −51

Drill 2: Functions Using Standard Notation

14 D

15 A PITA for n, plug each answer choice into h(x), and see which

one spits out 10 Alternately, you could solve 10 = n2 + n − 2 by setting n equal to 0 and factoring; the solutions are −4 and 3.

19 E The greatest factor of 75 not equal to 75 is 25 Therefore, f(75) =

75 • 25 = 1,875

20 E If y = 3, then g(−y) = g(−3) Because −3 < 0, g(−3) = 2|−3| =

2(3) = 6

37 E First, determine which part of the split function you need to use

in order to find f(2,3) and f(0.5,4) In both cases xy equals an

even number, so you only need to deal with the first part of the split function Then, simply plug in the given values:

f(2,3) + f(0.5,4) = 9 + 4 = 13

Choose (E)

Drill 3: Compound Functions

2 D Plug In a number for x Try 3 f(g(3)) = f(7) = 21, and g(f(3)) =

g(9) = 13, so the difference is 8.

8 E To evaluate f(g(−2)), first find the value of g(−2), which equals

(−2)3 − 5, or −13 Then put that result into f(x): f(−13) =|−13| −

5 = 13 − 5 = 8

9 B Let’s PITA Plug In 3 for g(x): f(3) = 5 + 3(3) = 14 Nope—

eliminate (A) Now let’s Plug In 4 for g(x): f(4) = 5 + 3(4) = 17.

Any of the other choices would leave a variable in the

16 D Just Plug In a nice little number, perhaps x = 3 You get g(f(3))

= g(64) = 12 Now just plug 3 into the answers for x, to see

which one hits your target number, 12

1.804

22 E Plug In The Answers! You are looking for the choice that makes

eliminate (A), (B), and (C) Next, try (D):

; also too small, so eliminate (D) and pick (E)

35 B Plug In! Make x = 3, so Next, plug x = 3

into each answer choice:

(A)

(B)

(C)

(D)

(E) 2 − (3) = − 1

Finally, plug each answer into f(x) and look for the answer that

equals − Only (B) works

7 B Plug a number into f(x) For example, f(2) = 1.5 Since g(1.5) =

2, the correct answer is the function that turns 1.5 back into 2 Choice (B) does the trick

18 B PITA, starting with (C) Take each answer choice, plug it in for x

in f(x), and see which one spits out 9.

30 E The fact that f(3) = 9 doesn’t tell you what f(x) is It’s possible

that f(x) = x2, or that f(x) = 3x, or that f(x) = 2x + 3, and so on.

Each of these functions would have a different inverse function The definition of the inverse function cannot be determined

31 C Plug In! This question is looking for the inverse of f(x) Make x

= 10, so Next, plug 2 into each answer choice and look for an answer that equals 10 Only (C) fits

Drill 5: Domain and Range

9 A This function factors to Three values of x

will make this fraction undefined: −2, 0, and 3 The function’s domain must exclude these values

15 E This function factors to The product of

these binomials must be nonnegative (that means positive or zero), since a square root of a negative number is not a real number The product will be nonnegative when both binomials

are not positive (x ≤ −2) or when both are nonnegative (x ≥ 6) The function’s domain is {x: x ≤ −2 or x ≥ 6}.

16 D Take this one step at a time Because a number raised to an

even power can’t be negative, the range of a2 is the set of

nonnegative numbers—that is, {y: y ≥ 0} The range of a2 + 5 is

found by simply adding 5 to the range of a2, {y: y ≥ 5} Finally,

to find the range of , divide the range of a2 + 5 by 3,

, or {y: y ≥ 1.67} The correct answer is (D).

19 D Because this is a linear function (without exponents), you can

find its range over the given interval by Plugging In the bounds

of the domain f(−1) = −1, and f(4) = 19 Therefore the range of f

is {y: −1 ≤ y ≤ 19}.

28 E Start by finding the smallest value for f(x) within the domain −3

≤ x ≤ 3 Because x4 is a variable to an even exponent, the

smallest value for that term is 0 If x4 = 0, then the fraction becomes , which is the lowest part of the

range of f(x); eliminate (A) and (C) Next, check the extremes of

Checking values between −3 and 3

confirms that 26.333 is the greatest value for f(x); choose (E).

Drill 6: Identifying Graphs of Functions

1 D It’s possible to intersect the graph shown in (D) twice with a

vertical line, where the point duplicates an x-value on the curve.

3 B It’s possible to intersect the graph shown in (B) more than once

with a vertical line, at each point where the graph becomes vertical

Drill 7: Range and Domain in Graphs

17 A The graph has a vertical asymptote at x = 0, so 0 must be

excluded from the domain of f.

24 D Only two x-values are absent from the graph, x = 2 and x = −2.

The domain must exclude these values This can be written as

{x: x ≠ −2, 2} or {x: x ≠ 2}.

28 C The graph extends upward forever, but never goes lower than

−3 Its range is therefore {y: y ≥ −3}.

37 C Plug In a big number, such as x = 1,000 It looks like y

approaches 5

48 E Plug the numbers you are given into the equation to see what

happens to the graph In I, if x = 2, then y = − , which does not

exist Therefore I is definitely an asymptote, and you should eliminate answer choices without I in them, that is, (B) and (C)

Now, try Plugging In a big number for x, like x = 1,000 y heads toward −1, which means y = −1 is also an asymptote, and III is

correct Cross off answer choices without III in them, in other words, (A) and (D) The correct answer is (E)

Drill 8: Roots of Functions

16 D PITA! Plug In each choice for x into f(x) to see which one spits

out 0

19 C The function g(x) can be factored as g(x) = x(x + 3)(x − 2) Set

this function equal to zero and solve for x You’ll find the

function has three distinct roots, −3, 0, and 2

25 D The roots of a function are the x-values at which f(x) = 0 In

short, the roots are the x-intercepts—in this case, −4, −1, and 2.

Drill 9: Symmetry in Functions

6 D “Symmetrical with respect to the x-axis” means reflected as

though the x-axis were a mirror That is, the values of the function above the x-axis should match corresponding values below the x-axis.

17 E An even function is one for which f(x) = f(−x) This is true by

definition of an absolute value Confirm by Plugging In numbers

30 A For an odd function, all points are reflected across the origin

Therefore, every point in quadrant I will have a reflection in quadrant III, and every point in quadrant II will have a reflection in quadrant IV The only graph which satisfies this requirement is (A)

34 C

If f(x) is a function, it has no more than one y-value for any x- value; therefore, it cannot be symmetrical with respect to the x-axis Eliminate (D) and (E) For any even function, f(−x) = f(x),

so plug some numbers in Try x = 2 and x = −2 If x = 2, then f(2) = 0.296, and if x = −2, then f(−2) = −0.296 These are not equal; eliminate (B) If a function is odd, then f(−x) = −f(x) In this case, we can see that f(−2) = −f(2) You can plug in other values to confirm, but each choice will satisfy f(−x) = −f(x), so

the function is odd Choose (C)

Drill 10: Degrees of Functions

8 B Count the number of times the graph crosses the x-axis Each

intersection is a distinct real root

17 E The graph shown has five visible distinct x-intercepts (zeros), so

it must be at least fifth-degree The degree of a function is determined by its greatest exponent Only the function in (E) is

at least a fifth-degree function

20 D Since the degree of a function is determined by its greatest

exponent, all you need to do in order to find the fourth-degree function is figure out the greatest exponent in each answer choice when it’s multiplied out Remember, you don’t need to

do all of the algebra; just see what the greatest exponent will be Choice (A) is a second-degree function, because its

highest-order term is x2 Choices (B) and (C) are third-degree functions,

because the highest-order term in each function is x3 Choice

(E) is a fifth-degree function, since x • x • x3 = x5 Only (D) is a fourth-degree function

Comprehensive Functions Drill

1 B Start by putting 3 and 1.5 into the function for x and y:

(3)2(1.5)+2(1.5)−(1.5) = 21 Next, plug each answer choice into

the function, using the first term for x and the second term for

y The only one that equals 21 is (B).

3 C Start with the inside function: Then

plug this value into f(x): , (C)

5 E Because the numerator is an odd root, it does not affect the

domain of g(x) The restrictions on the domain are entirely due

to the denominator You cannot have a denominator of 0;

therefore, x cannot equal 3 Eliminate (B), (C), and (D) You cannot take the square root of a negative number, so x cannot

be less than 3 either Eliminate (A) and choose (E)

11 E For a function to be even, it must be reflected across the y-axis.

This function does not fit that description; points in the first quadrant do not have their corresponding points in the second

Eliminate (A) and (D) An odd function has symmetry across

the origin The algebraic definition is that f(−x) = −f(x) for all

points in the domain This is the case for this graph, so II is

true; eliminate (C) Finally, if you draw the line y = −x, you find

that the function is symmetrical across this line III is also true,

so choose (E)

18 D The function g(x) is an upward-opening parabola with its vertex

on the origin f(x) opens downward, so you need a function with

a −g(x); eliminate (A) and (B) f(x) is shifted to the left of the y-axis, so you need to add within the parentheses; eliminate (E)

Finally, f(x) has its vertex above the x-axis, so you need to add

outside the parentheses; eliminate (C) and choose (D)

27 C To have an asymptote, a function must be a fraction Eliminate

(B) and (E) A function has a vertical asymptote wherever its

denominator is equal to 0 Plug x = 3 into the denominator of each function The only denominator that equals 0 when x = 3

is (C)

28 D To find f(f(f(−2))) start with the innermost term Because −2 <

−1, use the uppermost definition for the first term: f(−2)=

(−2)2−3 = 1 Now, find f(f(1)) 1 falls into the middle range, so apply that definition: f(1) = e1 = 2.718 Finally, find f(2.718) by using the last definition: f(2.718), (D).

32 A Plug In! Make x = 4, so Plug 0.125 into each

answer choice and look for the answer that equals 4 Only (A) works

44 B Plug In! Make x = 3, so f(g(3)) = 2(3)2 = 18 and g(3) = (3)2 − 1 =

8 Plug 8 into each answer choice and look for the answer that equals 18 Only (B) works

45 D In (A), taking the absolute value of a function will reflect any

negative portion of the function across the x-axis, and the

negative absolute value will reflect the result back across the x-axis However, the function will remain periodic Choice (B) moves the whole function left two units and up two units, but the function will remain periodic Choice (C) stretches the function vertically by a multiple of 3 and shifts the graph to the right 2 units, but it still is periodic In (D), you find a different

kind of transformation Multiplying f(x) by x will change the value of the function in a non-repeating manner because x

keeps changing (as opposed to the other answer choices in

which the transformation is constant) For example, take f(x) = cosx The period is 2π, so f(0) = f(2π) = 1 However, if you multiply f(x) by x, the function is now xf(x), so but (2π) f(2π) = (2π) cos (2π) = 2π Other values of x will also show that xf(x) is

no longer periodic

If f(x) is periodic, (E) will also be periodic, as you are repeatedly dividing 1 by the same values of f(x).

49 D If f(g(x)) = x, then g(x) is the inverse of f(x) To find the inverse,

choose points on the graph of f(x) and eliminate answer choices that do not have a corresponding point Estimate that the y-intercept of f(x) is (0, 2) If f(0) = 2, then g(2) = 0, so you need a

graph that includes the point (2, 0) Eliminate (A), (B), and (E)

Next, estimate that x-intercept of f(x) is (−2, 0) This means g(x) must include the point (0, −2) Eliminate (C) and choose

(D)