Algebra II Through Competitions Chapter 7 Function Composition and Operations
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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.
Algebra II Through Competitions Chapter 7 Function Composition and Operations
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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.
Algebra II Through Competitions Chapter 7 Function Composition and Operations
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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.
Algebra II Through Competitions Chapter 7 Function Composition and Operations
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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).
Algebra II Through Competitions Chapter 7 Function Composition and Operations
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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?
Algebra II Through Competitions Chapter 7 Function Composition and Operations
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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)
Algebra II Through Competitions Chapter 7 Function Composition and Operations
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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
Algebra II Through Competitions Chapter 7 Function Composition and Operations
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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.
Algebra II Through Competitions Chapter 7 Function Composition and Operations
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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
Algebra II Through Competitions Chapter 7 Function Composition and Operations
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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
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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 ).
Algebra II Through Competitions Chapter 7 Function Composition and Operations
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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.
Problem 3. Solution: E.
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 .
Problem 5. Solution: D.
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 .
Algebra II Through Competitions Chapter 7 Function Composition and Operations
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Problem 6. Solution: E.
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 .
Problem 8. Solution: D.
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.
Problem 9. Solution: A.
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.
Problem 10. Solution: A.
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 .
Problem 11. Solution: A.
12 = (x + 5)2
+ 8 4= (x+5)2
±2 = (x + 5), so x = −3 or x = −7 .
So −3 + (−7) = −10.
Problem 12. Solution: D.
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
Algebra II Through Competitions Chapter 7 Function Composition and Operations
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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.
Problem 14. Solution: A.
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.
Problem 15. Solution: B.
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.
Problem 16. Solution: E.
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.
Problem 17. Solution: A.
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.
Problem 18. Solution: C.
In fact, f(n) = 2n + 3, and T = f(f(f(f(5)))) = f(f(f(13))) = f(f(29)) = f(61) = 125.
Problem 19. Solution: B.
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.
Problem 20. Solution: A.
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.
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Problem 21. Solution: C.
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 .
Problem 23. Solution: D.
For odd function, we have )()( xfxfy . We test and we know that D is the answer.
Problem 24. Solution: B.
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.
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