A2A MAT115 Practice Exam 3 Answers
An equation and its graph are given.
A) Use the graph to determine whether the graph is symmetric with respect to y-axis(even), x-axis, origin(odd), or it has
no symmetry.
B) Determine algebraically if functions 1-4 are even, odd or neither. Your answer should match what you deduce from
the graph.
Algebraic test model: ODD:  f ( x)  f ( x) EVEN: f ( x)  f ( x)
1.
f ( x)  x 4  2 x 2  1
Odd test:
 f (x)
Even function (from graph and algebraically)
?
f ( x)
Even test:
?
 f ( x)   ( x 4  2 x 2  1)

?
f ( x)  x 4  2 x 2  1
y
?
?
f (  x)  x 4  2 x 2  1
f ( x)
f (x )
Function is not odd
2.
f ( x)
f ( x)  ( x) 4  2( x) 2  1
f (  x)  x 4  2 x 2  1
 f (x)
?
f ( x)  x 4  2 x 2  1
f ( x)  ( x) 4  2( x) 2  1
 f ( x)   x 4  2 x 2  1
f (x )

f ( x)
Function is even
3
x 3
2
4.
x  y2
Even function (from graph
3.
g x   x3  x
Neither (from graph)
Odd function (from graph)
5. What is the domain and range of the above graphs? Write your answer in interval notation (set notation is also
shown)
Graph 1: Domain:  ,  Range: 0.
Graph 2: Domain: [-2, 2] Range: [0, 3]
OR
Domain : {x| x   (all real numbers)}
Graph 3: Domain: 0, 
Domain : {x| x  0}
Range:
OR
Range: {y| y  0}
Domain : {x| x
 , 
 2  x  2}
Range: {y|
Graph 4: Domain:  , 
OR
Range: {y| y   }
Range:
OR
Domain : {x| x
Range: {y| y
 , 
 }
f ( x)  x  4
6. Rewrite the equation 2x + 3y = 5y – 8 using function notation
7. Complete the following table
 }
 0  x  3}
x
f x   3x  1
-2
-1
0
1
2
27
12
3
0
3
3
12
2
8. A farmer has a rectangular garden plot surrounded by 344 feet of fencing. Find the dimensions of the garden if its
area is 7227 ft2.
73 feet by 99 feet
9. The graph to the right is a graph of h(x).
a.
h(3) = -3 h(-1) = 5 if h(x) = 4 x = 0 if h(x) = -3 x = 1, 3
b. Find the domain and range of h(x)
Domain: 𝑥|𝑥 = −4, −3, −1, 0, 1, 3
Range: 𝑦|𝑦 = −4, −3, 3, 4, 5
10. For each of the following representations of a relation, state whether the relation is or is not a function. Explain your
answer.
h(x) = function because the x’s don’t repeat
x
-2
-1
0
1
2
s(t) = not a function because x’s repeat
y
3
2
1
2
3
(-2, -3), (-1, -2), (0, -1), (-1, 0), (-2,1)
11. What is the domain and range of h(x), f(x) and g(x) below? Write your answers in interval notation and in set
notation.
f(x)
g(x)
h(x)
hy = y2-2
fx = x3+2
4
4
2
2
-5
5
-5
5
-2
-2
-4
-4
D: {𝑥|𝑥 ≥ 1} R: {y| y   }
D : {x| x   } R: {y| y   }
D: [−2, ∞) R: (−∞, ∞)
D: (−∞, ∞)
D: {𝑥| − 3 ≤ 𝑥 < 2}
R: {𝑦| − 3 ≤ 𝑦 < 6}
D: [−3, 2)
R: (−∞, ∞)
R: [−3, 6)
12. Using the above graphs of f(x), g(x) and h(x) above
a.
f(-2) = 0; f(2) = 2 and -2, g(0) = 2, g(1) = 3, h(-3) = 1, h(-1) = -3, h(1) = 1
b.
g(x) = 2 then x = 0, if f(x) = -2 then x = 2, if f(x) = 0 then x = -2,
g(x) = 1 then x = -1, if h(x) = -2 then x = -2 and 0
13. Given the function g ( x)  3x 2  4 x
a. Find
g ( x  h)  g ( x )

h
6x + 3h – 4
b. Find g ( x  2 )  3x 2  2 2 x  4x  4 2  2
TRANSFORMATIONS
Know the basic graphs of the following:
14.
a.
yx
d.
y
b.
y  x2
c.
f. y 
e. y  x
x
a. f  x   3 x  4  5
y  x3
b.
1
x
f x   3  2( x  1)2
c.
f x   4  3  x
15. Given the graph on the right find the equation for base graph f(x) (blue) and the
transformations applied to create g(x) (maroon).
y  2 x  4  2
16. Given y  2 f ( x  1)  4 , describe in words how the graph of f(x) has been transformed to create y.
y = f(x + 1)

Shift horizontally left 1 (shift against the sign)
y = 2f(x + 2)

Double y-values by 2
y = 2f(x + 2) + 4 
Shift vertically up 4 (shift with the sign)
17. Given the graph of h(x) below, draw y  2 f ( x  3)  2
The order of transformations is critical and must be a follows:
Horizontal shift left or right  reflection across y-axis  reflection across x-axis  stretching factor of yunits  vertical shift up or down.
1.
2.
3.
4.
2 f (–x + 3) – 2
2 f (–x + 3) – 2
2 f (–x + 3) – 2
2 f (–x + 3) – 2
Shift horizontally left 3
Reflect across the y-axis
Multiply the y-values by 2
Shift down horizontally 2
18. The collision impact of an automobile varies jointly as its mass and the square of its speed.
Suppose a 2000lb car traveling at 55 miles per hour has a collision impact of 6.1. What is the collision
impact of the same car at 65 miles per hour?
C  kms2

k
6 .1
200055
 C
2
6.1
200055
2
ms 2 C = 8.52
19. a) Given the piecewise graph of f(x) to the right, find f(-2) and f(2)
f(-2) = -2.
f(2) = 6.
b) Find the domain of f(x)
 ,6 .
c) Determine the range of f(x) in interval notation.
(-4, 6).
d) Describe f(x) as a function.
x  2
  2 if
 x 2  2
if  2  x  2
f ( x)  
3
 x  7 if x  2
 2
20. Given the piecewise function g (x ) below, find g(0), g(-5), g(2), g(3), then
graph g(x) using transformations and the domain restrictions.
x  2
  3 if

x

2

if  2  x  3
g ( x)  
2
 x  4 if
x3

3
g(0) = |0| – 2 = – 2
g(-5) = – 3
g(2) = |2| – 2 = 0
g(3) =
2
(3) + 4 = 7
3
21. Find the domain and range of the following functions :
f ( x)  2 x 3  3 x 2  5
g ( x)  2 x  1
f ( x)  2 x 3  3 x 2  5
Domain:
Range:
h( x ) 
2x
x 4
2
m( x ) 
x4
2x  3
 ,  This problem has no fractions or radicals.
 ,  Look at a basic cubic graph. It goes up and down forever.
Domain: x 
g ( x)  2 x  1
Range
h( x ) 
2x
x 4
x | x  , x   2
If you graph the equation, you will see the range is  ,  .
Domain: x 2  4  0 so the domain is
2
Range:
m( x ) 
0, 
1
1 
so the domain is  ,  
2
2 
x4
2x  3


3
3
 First you have 2x  3  0 so x  
2
2
You also the radical so x  4  0 which gives  4, 
Range: If you graph the equation, you will see the range is  ,  .
Domain:  x | x  4, x  
4
x 4
h( x ) 
g ( x)  1  x
22. Use the following functions to answer a – j.
f ( x)  x 2  2 x
2
a. Determine the domain of each function individually
g(x): domain is [1, ) , h(x): domain is x | x  , x   2. f(x): domain is  ,  .
b. Find (f + h)(x) and list the resultant domain in set notation
f
 hx  = x 2  2 x 
4
x 4
Domain: { x x  , x  2,2 }
2
c. Find (g/f)(x) and list the resultant domain in set notation
g / f x
=
1 x
x  2x
Domain: { x x  1; x  0 }
2
d. Find ( f  h)( x) and list the resultant domain in set notation
 f  hx
4 x2  8x
=
x2  4
e. Find ( f  h)( 3) =
Domain: { x x  , x  2,2 }
f(-3) – h(-3) = 3 –
 1 x  2
2
4
=
5
11
5
1  x = 1 x  2 1 x
f.
f ( g ( x)) or ( f  g )( x) =
g.
h( f (1)) or (h  f )( 1) =
h.
g ( f (2)) = f 2   2   22  = 8  g 8  1  8
i.
f ( x  3) * 2 f ( x)  x  3  2x  3 * 2 f x  = x 2  4 x  3 * 2 f x  = x 2  4 x  3 * 2 x 2  2 x

4
3
2
=
9 = 3

2
= 2 x 4  4 x 3  10 x 2  12 x
23. Given f(x) is in maroon and h(x) is in blue, find the following:
a.
( f  h)( 2) = -1
c.
( f * h)( 3) = -4
b.
( f  h)( 6) = 3
d.
( f  f )( 4) = -4
 

24. a) Given f ( x) 
f
1
3x  2
find f-1(x). State the domain and range of f(x) and f-1(x).
4
x   4 x  2 .
3
b) Given T ( x) 
T
1
The domain and range for both is  , 
2 x  6 , find f-1(x). State the domain and range of T(x) and T-1(x).
2

x  6
x  
2
. The domain of f(x) to be 0,   . The range is 6,  
Hence the domain of T 1 x  is 6,   and the range is 0,   .
25. Use the inverse function property (composition) to determine if f ( x)  5 x  2 and g ( x) 
1
2
x  are inverses.
5
5
The inverse property function (composition) should show that f ( g ( x)) = x and g ( f ( x)) = x when simplified.
26. Find the inverse functions for the following functions. Write your answer in terms of f 1 x  .
a.
f x  
3x  2
2
so f 1  x  
5x
5x  3
b. f x   2 x 2  4 ; x > 1 so f 1 x  
x4
; x > -2
2
27. Given the graph S(t) on the right:
a. State the domain and range or S (t ) in set notation.
Domain: { x x  4,3,0,1,2,4 }
Range: { y y  3,1,3,4,5 }
b. Is S(t) a one-to-one function? Why.
No. One-to-one means each x-value is only used once and each y-value is only used once. The y-value of 4 is
used twice at (0, 4) and at (2, 4). It is a function but it fails the horizontal line test so it is not 1-to-1.
c. Explain in words how the domain and range of S(t) is related to the domain and range of
its inverse. The domain and range of f and its inverse are exact opposites.
d. Draw the inverse function for S(t). Is the graph one-to-one?
I take each point from the original and switch the x and y-values.
The graph is not one-to-one since the original graph was not 1-to-1 and it does not
pass the horizontal line test.
28. Be able to write your answers in set notation or interval notation depending on the directions and know the
difference.
Example:
(, 2]
 {x | x  2}
(, 4)  (4, )
(24, 15)

x | x  , x  5
 {x | x  R,  24  x  15}