algebra ii through competitions chapter 7 function ... · algebra ii through competitions chapter 7...

Algebra II Through Competitions Chapter 7 Function Composition and Operations

1

1. FUNCTIONS

1.1 Definition

A function is a relationship between the independent variable x and dependent variable y.

Each value of x corresponds exactly one value of y.

Note two different values of x can have the same value of y, but one value of x cannot

have two different values of y.

For example, the price of fruits in a store is a function of fruit kind. One pound of apple

(x1) and one pound of orange (x2) can have the same price $1.99 per pound (y).

But one pound of apple (x) cannot have two different prices at the same time (one price

tag says $1.99 per pound (y1) and price tag says $0.99 per pound (y2).

Example 1. (2009 Indiana Algebra II) Let

3 ,3

31 ,1

1 ,

)(

xx

xx

xx

xf . Then f(0) + f(2) +

f(4) is equal to

A. 0 B. 9 C.10 D. 21 E. None of these

Solution: C.

f(0) = 0.

f(2) = 2 + 1 = 3.

f(4) = 4 + 3 = 7.

f(0) + f(2) + f(4) = 0 + 3 + 7 = 10.

Example 2. (2009 Indiana Algebra II) If f(x2) = 4 x f(x + 2) + 3, what is the value of

f(4)?

A. 1 B. 7

3 C. 2 D.

3

7 E.

7

22

Solution: B.

f(22) = 4 2 f(2 + 2) + 3 f(4) = 4 2 f(2 + 2) + 3 f(4) = 4 2 f(4) + 3 f(4) =

−3/7.


2

Example 3. (2007 NC Algebra II) Let g(x) = x2 + b·x + c and g(2) = –6. Determine g(5).

A. –15 B. 2c – 9 C. –1.5c D. 2.5c – 10 E. –4c

Solution: C.

g(2) = –6 = (2)2 + b·2 + c –10 − c = 2b b = –5 − c/2.

g(5) = (5)2 + b·5 + c = 25 + 5(− 5 − c/2) + c = − 5c/2 + c = –1.5c.

Example 4. (2007 NC Algebra II ) If f (1) = 2 and f (n + 1) = (f (n))2, what is the value of

f (4) ?

A. 4. B. 16. C. 64. D. 256. E. 65,536.

Solution: D.

f(2) = f(1 + 1) = (f(1))2 = 4 and f(3) = f(2 + 1) = ( f(2))

2 =16 , and f(4) = f(3 + 1) = (f(3))

2

= 162 = 256.

1.2. Domain and range

A domain is all the possible values of x. A range is all the possible values of y.

Example 5. (2003 NC Algebra II) Determine the range of the function x

xxf

3)(

.

Solution: E.

Let y = F(x) = (x + 3)/x = 1 + 3/x x

y3

1

Solving for x we get: 1

3

yx .

We see that y can never equal 1, but can equal anything else, the range of this function is

{y∈Reals, y ≠ 1}.

1.3. Vertical line test

If each vertical line intersects a graph at no more than one point, the graph is the graph of

a function.


3

Figure (a) is the graph of y2 = x and it is not a function.

Figure (b) is the graph of y = x and it is a function.

(a) (b)

2. SYMMETRY

2.1. Properties

Symmetric with respect to

y-axis x-axis Origin

)()( xfxfy )()( xfxfy )()( xfxfy

Even Function Nothing Odd function

Example 6. (2004 NC Algebra II) A function f is even if for each x in the domain of f, f

(x) = f (−x). A function f is odd if for each x in the domain of f, f (−x) = − f (x). Which of

the following statement(s) is (are) true?

I. The product of two odd functions is odd.

II. The sum of two even functions is even.

III. The product of an even function and an odd function is odd.


4

IV. If f is any function and the function F is defined by F(x) = [f (x) + f (− x)]/2, then F is

even.

A. All statements are true. B. Only I, II and III are true. C. Only II, III and IV are true.

D. Only II and III are true. E. Only III and IV are true.

Solution: C.

First note that I is false since the product of two odd functions, like f(x) = x and

g(x) = x3 is cleary an even function. Next note that the next two are clearly true. The final

option, that 2

)()()(

xfxfxF

is always even bears checking. First, if f(x) is even,

then 2

)()()(

xfxfxF

= )(

2

)()(xf

xfxf

, which is even. If f(x) is odd, then

2

)()()(

xfxfxF

= 0

2

)()(

xfxf, which is a constant function, hence it is also

even.

2.2 Graph of functions

To graph Shift )(xfy by c

cxfy )( Upward

cxfy )( Downward

)( cxfy Left

)( cxfy Right

Example 7. (2005 NC Algebra II ) Suppose we are given the graph of y = f (x). Which

functional expression below describes a graph that is reflected across the x-axis then

shifted 3 units to the right and 5 units up?

A. 3 − f (x − 5) B. f (x + 3) − 5 C. f (3 − x) + 5 D. f (x − 3) + 5 E. 5 − f (x − 3).


5

Solution: E.

When a graph is flipped over the x-axis, f (x) turns into − f (x). To move a graph to the

right, we must subtract from the x-value. To move a graph up we add to the

final value, so the desired formula is −f(x − 3) + 5 = 5 − f(x − 3) .

Example 8. (2004 NC Algebra II) If (1, 5) is a point of the graph of y = f (x), then the

graph of y = f (x − 3) contains the point (2, c) where c equals

A. 2 B. 3 C. 4 D. 5 E. 6

Solution: D.

If (1, 5) is a point on the graph of y = f (x) then the point (2, 5) will be the corresponding

point on the graph of y = f(x – 3), since this graph is the result of sliding the original

graph to the right 3 units.

3. OPERATIONS ON FUNCTIONS

Given two functions f(x) and g(x), we have the following operations

Sum: (f + g) (x) = f(x) + g(x)

Difference: (f g) (x) = f(x) g(x)

Product: (f g) (x) = f(x) ∙ g(x)

Quotient: (f/g) (x) = f(x)/g(x) g(x) ≠ 0.

The domains of f + g, f g, f g) include all real numbers of the intersection of the

domains of f and g.

The domain of f/g includes all real numbers of the intersection of the domains of f and g

for which g(x) ≠ 0.

4. COMPOSITE FUNCTIONS

The composite function of f and g) is (f ◦g)(x) = f(x) ◦g(x) = f (g(x)).

The domain of f ◦g is the set of all numbers x in the domain g such that g(x) is in the

domain of f.

Example 9. (2007 NC Algebra II) For f (x) = x2 −1 and g(x) = 2x + 3, which has the

greatest value?


6

A. f(g(3)). B. g(f (2)). C. g(f (−1)). D. f(g(−5)). E. g(10) .

Solution: A.

32)( xfxgf = 19124132 22 xxx and

123)1(2)1()( 222 xxxgxfg . So f(g(3)) = 80, g(f(2)) = 9, g(f(1)) = 3,

f(g(5)) = 48, and g(10) = 23, so f(g(3)) has the largest value.

Example 10. (1999 NC Algebra II ) Let be an operation defined on functions such

that: f g(x) f (g(x)) g(f (x)). If f(x) = x2

1 and g(x) = 2x + 1, find f g(x) .

A. x2

2x 2 B. 2x2

4x + 1 C. 2x2

+ 4x – 2 D. 2x2

+ 4x + 1 E. none of the above.

Solution: (D).

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

2 1) + 1) = 4x

2 + 4x + 1 1 (2x

2 2 + 1) = 2x

2 + 4x + 1.

5. INVERSE FUNCTIONS

5.1 One-to-one function

Horizontal line test: If each horizontal line intersects the graph of a function in no more

than one point, the function is one-to-one.

Figure (a) is the graph of y2 = r

2 x

2 and it is not one-to-one.

Figure (b) is the graph of y = 22 xr and it is one-to-one.

Figure (a) Figure (b)


7

5.2 Inverse function

f is a one-to-one function. g is the inverse function of f if (f ◦g)(x) = x for every x in the

domain of g and (g ◦f)(x) = x for every x in the domain of f.

g is written as f 1

.

Notes: (1) Do not confuse 1 in f 1

with a negative exponent.

(2) A function f has the inverse function if and only if f is one-to-one.

(3) The graph of f and f 1

are symmetric with respect to the line y = x, that is, the graph

of f 1

is the image obtained by reflecting the graph of f about the line y = x.

5.3. Steps in find an equation for f 1

(1) Check if f defined by y = f(x) is a one-to-one function.

(2) Solve for x. Let x = f 1

(y)

(3) Exchange x and y to get x = f 1

(y).

(4) Check that (f ◦ f 1

)(x) = x and (f 1

◦f)(x) = x.

Note: If it is hard to explicitly solve for x, we just exchange x and y to get an expression

that define the inverse function implicitly.

Example 11. (2000 NC Algebra II) Suppose f and g are both linear functions, with y =

2x + 1 and f(g(x)) = x. Find the sum of the slope and the y-intercept of g.

A. -2 B. −1 C. 0 D. 1 E. none of these

Solution: C.

If f(g(x)) = x, then f(x) and g(x) are inverse functions. To find the inverse of y = 2x + 1,

switch x and y and solve for y. The result is that y = g(x) = (1/2)x + 1. The sum of the

slope and y-intercept is 1/2 + 1/2 = 0

Example 12. (2003 NC Algebra II) Find, if possible, the inverse of 3

2)(

x

xxf .

A. 1

26

x

x, x 1 B.

2

23

x

x, x 2 C.

1

23

x

x, x 1 D.

2

26

x

x, x 2

E. no inverse exists


8

Solution: C.

To find the inverse, exchange x and y in the equation 3

2

x

xy and solve for y. Solving

for y, xy – 3x = y + 2 xy – y = 3x + 2 y(x – 1) = 3x + 3 3

2

x

xy , x 1 or

1

23)(1

x

xxf , x 1.

Example 13. (2005 NC Algebra II) If 32)1

1(1

xx

f then

A. f(x) = 2x – 4. B. f(x) = 2x – 5. C. 2

5)(

xxf . D.

22

2)(

x

xxf .

E. 5

2)(

xxf .

Solution: E.

If 321

11

xx

f , then 1

1)32(

xxf . Now let u = 2x – 3, so

2

3

ux

and 5

2

2

5

1

12

3

1)(

uuu

uf , or 5

2)(

xxf .

Example 14. (2009 Indiana Algebra II) If f(x) is a linear function and the slop of y = f(x)

is 2

1, what is the slope of y = f

1(x)?

A. 2 B. 2

1 C.

2

1 D. 2 E. None of these

Solution: D.

Let y = f(x) = bx 2

1 2y = bx 2 byx 22

bxxf 22)(1 .

The slope is 2.


9

PROBLEMS

Problem 1. (1999 NC Algebra II) Given: 1

1)(

t

ttf , evaluate

1

11

cf .

A. 1 / (1 + 2c) B. 1 / (2 + c) C. 2c – 1 D. c – 2 E. none of these

Problem 2. (1999 NC Algebra II ) Given f(x) = x3

+ x 5, find t so that f 1

(t) = 0.2

A. 4.792 B. 3 8.4 C. 2 D. 1.5437 E. none of these

Problem 3. (2000 NC Algebra II) If x

xf

2

2)2( for all x > 0, then 2f(x) =

A. x1

2 B.

x2

2 C.

x1

4 D.

x2

4 E.

x4

8

Problem 4. (2001 NC Algebra II ) Let 3

4)(3)1(

xhxh for x 1, 2, 3,… and ,

3

2)1( h , find h(3) .

A. 1 B. 0 C. 5 D. 2 E. 6

Problem 5. (2001 NC Algebra II) Given 1

1)(

x

xxf . Solve 3

11

xf .

A. x = 1 B. x = 2 C. -1/2 D. 2

1x E.

3

1x

Problem 6. (2003 NC Algebra II ) If f (x) f (x 2) x, and f (7) 11, find f (5).

A. 10 B. 8 C. 6 D. 8 E. 4

Problem 7. (2003 NC Algebra II) Given 4

1)( xxf and ,)( 2

3

xxh find (h f1

)( 2

1).

A. 2

1 B.

4

1 C.

8

1 D.

4

1 E.

8

1

Problem 8. (2004 NC Algebra II) Let f be a linear function with the properties that f (1)

≤ f (2) , f (3) ≥ f (4) and f (5) = 5. Which of the following statements is true?

A. f (0) < 0 . B. f (0) = 0. C. f (1) < f (0) < f (−1) D. f (0) = 5 E. f (0) > 5


10

Problem 9. (2005 NC Algebra II) Let f ( x ) = x 2 −12 x + 5 , g ( x ) = x + a and f ( g ( x ))

= x 2 + c . Find the value of c.

A. –31 B. –7 C. a 2 −12 a D. 5 – 12a E. 5.

Problem 10. (2005 NC Algebra II) A function of the form bx

axf

)( has the

following properties f(1) = 3 and f 1

(5) = 1. What is the value of f(0)?

A. 4

15 B.

5

3 C. 4 D.

7

15 E. None of these

Problem 11. (2007 NC Algebra II) If f (x) =(x + 5)2 + 8, then what is the sum of the

values of x for which f (x) = 12?

A. -10. B. -7. C. 10. D. 20. E. 297.

Problem 12. (2007 NC Algebra II) Let f (x) be defined as the least integer greater than

x/5 . Let g(x) be defined as the greatest integer less than x/5 . What is the value of g(18) +

f (102)?

A. 21. B. 22. C. 23. D. 24. E. 25.

Problem 13. (2007 NC Algebra II) A function f from the integers to the integers is

defined as follows:

even is if 2/

odd is if 3)(

nn

nnnf .

Suppose k is odd and f (f (f (k))) = 27. What is the sum of the digits of k?

A. 3. B. 6. C. 9. D. 12. E. 15.

Problem 14. (2007 NC Algebra II) If f(1 – t 1

) = (5t + 1)/ t then f(t) = ?

A. 6 – t B. 1

16

t

t C.

14

1

t

t D.

t

15 E.

15

41

t

t

Problem 15. (2010 Indiana Algebra II) If f(x) = x2 – 3 and g(x) = 5 – x, then f(g

-1(x)) is

equal to:

A. x2 + 10x + 22 B. x

2 10x + 22 C. 35 x D. 35 x E. none of these

Problem 16. (2009 Indiana Algebra II) If f(x) = 1 – 4x and f 1

(x) is the inverse function

of f(x), then f(3) f 1

(3) is equal to

A. 1 B. 3 C. 4 D. 10 E. 13


11

Problem 17. (2008 Indiana Algebra II) If f(x) = 3x + 2 and g(x) = 3x + 1 then g(f 1

(x)) =

A. x – 1 B. x – 5 C. 9x + 3 D. 9x + 7 E. 9x + 5

Problem 18. (2010 UNC- Charlotte Algebra II) Suppose f(0) = 3 and f(n) = f(n − 1) + 2.

Let T = f(f(f(f(5)))). What is the sum of the digits of T?

A. 6 B. 7 C. 8 D. 9 E. 10.

Problem 19. (2004 UNC- Charlotte Algebra II) Let the function f be defined by f(x) = x2

+ 40. If m is a positive number such that f(2m) = 2f(m) which of the following is true?

A. 0 < m 4 B. 4 < m 8 C. 8 < m 12 D. 12 < m 16 E. 16 < m

Problem 20. (2011 Alabama Algebra II) Function f satisfies f(x) + 2f(5 − x) = x for all

real numbers x. The value of f(1) is

A. 7/3 B. 3/7 C. 5/2 D. 2/5 E. None of these

Problem 21. (2010 Alabama Algebra II) Suppose that f(n + 1) = f(n) + f(n −1) for n = 1,

2, … . Given that f(6) = 2 and f(4) = 8, what is f(3) + f(5)?

A. −18 B. −19 C. −20 D. −21 E. −22

Problem 22. (1999 NC Algebra II) If f (x) x2 c

2, solve for x when (f ◦ f )(x) 0 .

A. x c B. x (c c ), x c c ) C. x cc 2 , x cc 2

D. x c2 c, x c

2 c E. none of the above.

Problem 23. (2008 Tennessee Algebra II) Which of the following functions is an odd

function?

A. 523)( 24 xxxf B. xxf )( C. cos)( f

D. xxxxf 352)( E. 22 1)( xxxf

Problem 24. Find the inverse of: 84)( 3 xxf .

A. 31 2)( xxf B. 31

4

8)(

x

xf C. 31

4

8)(

x

xf

D. The inverse does not exist E. None of these.

Problem 25. (2012 Illinois Algebra II) Let 3

189)2(

3

xxf . Then f

1(x) = kx

w + f

where k, w, and f are positive integers. Find the value of (2k + 3w+ 4 f ).


12

SOLUTIONS:

Problem 1. Solution: E.

Let 1

1

t

ty . If the y's and the t's are switched, then solving for y arrives at the inverse.

1

1

y

yt

t

tytf

1

1)(1

1

11

cf =

c

c1

21

= 1 2c.

Problem 2. Solution: A.

The input of f(x) will be the output of f1

(x) and the output of f(x) will become the input of

f 1

. This means that x = f 1

(t) = 0.2 and t = f(x) = f(0.2) = 4.792.


The x in the denominator of f(2x) = 2/(2+x) represents half of the function’s input. So f(x)

= 2/(2+1/2 x) = 4/(4+x) . 2f(x) = 8/(4+x).

Problem 4. Solution: D.

3

4)(

3

4)(3)1(

xh

xhxh , so each term is just four-thirds greater than the one

before it. So 3

2)1( h ,

3

2)2( h +

3

2

3

4 , and

3

2)3(h 2

3

4 .


To find the inverse, exchange the x and y and solve for y. Thus 1

1

y

yx xy – x = y +

1 xy – y = x + 1 1

1

x

xy . From this we see that the fUnction is its own inverse. If

we had graphed it, and noticed the symmetry about the line y = x, we could have drawn

the same conclusion. Thus 1

1

11

11

111

x

x

x

x

xf

xf . Setting this equal to 3,

we have 1

1

x

x=3 x +1 = 3(x – 1) 2x = 4, so

2

1x .


13


Since f(x) = f(x − 2) + x, f (7) = f (5) + 7 ⇒ 11 = f (5) + 7 ⇒ f (5) = 4.

Problem 7. Solution: C.

To find f1

, 4

1 yx 4x = 4y + 1

4

14

xy f

1(x) = x

4

1 . Then

(h f 1

)( 2

1) =

8

1

4

1

4

1

4

1

2

1

2

12/3

1

hhfh .


Since f is a linear function and f(1) f(2), we know that the slope is greater than or equal

to zero. Likewise, since f(3) f(4), the slope is less than or equal to zero. Taken together,

this means that the slope must be zero, making the function f(x) = 0x + 5, so f(0) = 5.


f(g(x))= f(x + a)=(x + a)2−12(x + a) + 5=x

2+ (2a −12)x+(a

2 − 12a + 5).

Since we are told that f (g(x))= x2 +c, we know that 2a−12=0 and a

2−12a+ 5 = c ,

so a = 6 and a2

−12a+5=36 – 72 + 5= −31 = c.


b

af

1)1( = 3 and 5

1)1(1)5(1

b

aff , so a = 3 + 3b and a = – 5 + 5b and

3 + 3b = – 5 + 5b 8 = 2b b = 4 and a = 3 + 3(4) = 15, making 40

15)0(

f .


12 = (x + 5)2

+ 8 4= (x+5)2

±2 = (x + 5), so x = −3 or x = −7 .

So −3 + (−7) = −10.


g(18) = 4 , since 4 is the least integer greater than 18/5 . Similarly f (102) = 20, since 20

is the greatest integer less than 102/5. So g(18) + g(20) = 4 + 20 = 24.

Problem 13. Solution: B.

Since k is odd, f(k) = k + 3. Since k + 3 is even, f(k + 3) = f(f(k)) = 2

3k. If

2

3k is odd,

then 27 = f(f(f(k))) =

2

3kf =

2

3k + 3, which imlies that k = 45. This is not possible


14

because f((f(45))) = f(f(48)) = f(24) = 12. Hence 2

3k must be even, and

27 = f(f(f(k))) =

2

3kf =

4

3k, which implies that k = 105. Checking, we find that

f((f(105))) = f(f(108)) = f(54) = 27. Hence the sum of the digits of k is 1 + 0 + 5 = 6.


Let x = 1 – t –1

then t –1

= 1 – x.

f(1 – t –1

) = 5 + t –1

f(x) = 5 + (1–x) = 6 – x. f (t ) = 6 – t.


g-1

(x) = −x + 5.

We know that If f(x) = x2 – 3, so f(g

-1(x)) = (g

-1(x))

2 – 3 = = (–x + 5)

2 – 3 = x

2 + 25 – 10x

– 3 = x2 10x + 22.


Since f(x) = 1 – 4x, f(3) = 1 – 4 (−3) = 13.

Since f 1

(x) = 4

1

4

1 x , f

1(−3) =

4

1)3(

4

1 = 1.

f(3) f 1

(3) = 13 1 = 13.


Since f(x) = 3x + 2, f 1

(x) = 3

2

3

1x .

Since g(x) = 3x + 1, g(f 1

(x)) = 3 f 1

(x) + 1 = 3 (3

2

3

1x ) + 1 = x – 2 + 1 = x – 1.


In fact, f(n) = 2n + 3, and T = f(f(f(f(5)))) = f(f(f(13))) = f(f(29)) = f(61) = 125.


Note that f(2m) = (2m)2 + 40 = 4m

2 + 40 and 2f(m) = 2(m

2 + 40) = 2m

2 + 80. It follows

that 2m2 = 40, so 4 < m 8.


We know that f(x) + 2f(5 − x) = x.

f(1) + 2f(5 − 1) = 1 (1)

f(4) + 2f(5 − 4) = 4 (2)

(2) 2 – (1): 3f(1) = 7 f(1) = 7/3.


15


f(6) = f(5) + f(4) = f(4) + f(3) + f(3) + f(2) = 5f(2) +3 f(1) = 2 (1)

f(4) = f(3) + f(2) = f(2) + f(1) + f(2) = 2 f(2) + f(1) = 8 (2)

(2) 3 – (1): f(2) = 22. Thus f(1) = 8 – 44 = – 36.

f(3) + f(5) = f(2) + f(1) + f(4) + f(3) = f(2) + f(1) + f(3) + f(2) + f(2) + f(1)

= 4f(2) + 3f(1) = 4 22 + 3 (– 36) = – 20.

Problem 22. Solution: (C).

f(f(x)) = f(x2 c

2) = (x

2 c

2)2 c

2 = 0

Let y = x2 and b = c

2. 0)(2 22 bbyby

2

)(4)2(2 22 bbbby

= bb .

222 ccx ccx 2 .


For odd function, we have )()( xfxfy . We test and we know that D is the answer.


Let f(x) = y.

84)( 3 xxf 84 3 xy 348 xy 3

4

8x

y

3

4

8

yx .

We switch the position of x and y: )(1 xf 3

4

8

x.

Problem 25. Solution: 31.

Let x + 2 = X.

3

189)2(

3

xxf

3

)4(9)(

3

XXf .

Let f(X) = y.

3

)4(9)(

3

XXf

3

)4(93

Xy 3 )4(93 Xy 43 3 yX .

We switch the position of X and y: 43)( 31 XXf .

So k = 3, w = 3 and f = 4. 2k + 3w+ 4 f = 6 + 9 + 16 = 31.

algebra ii through competitions chapter 7 function ... · algebra ii through competitions chapter 7...

Documents