1
Math 115 Practice Test 2
TOTAL =
37
Note: All working leading to a solution must be shown for full marks to be awarded!
Answers only will result in no marks being assigned.
f  x
f 
[3] 1. If f  x   5  x and g  x   x  3 find    x  
, and give its domain in interval notation.
g  x
g
[3] 2.If f  x   2x2  8x  4 and g  x   3x2  12x  5 , find the solution set for: f  x   g  x  . .
[3] 3. For the function
f  x 
[3] 4. Given the graph of
a
x
2
, find and simplify the difference quotient
y  f  x
f  x  h  f  x
h
. [3]
shown below, show and explain all transformations you use to obtain the
1
graph of y   f 2  x  2  3
2
   x  5  x  3

f  x   x  7
3  x  5 .

x5

3

2
[3] 5. Graph the piecewise relation defined by:
Identify the domain and range
and determine whether or not this relation is a function and justify your answer.
[3] 6. For the function whose equation is f  x  
5
x4
:
a) Find the average rate of change from ( 1, f(1)) to (2, f(2)).
b) The equation of the secant line passing through those two points.
[4] 7. Analyze the function
 1
 2 x  1

p  x   2 x  6

3 x  3

x  2
2  x  3 .
x3
Find the following:
a) Domain and range. b) Maximum and minimum values. c) The intervals where the function is increasing
or decreasing. d) The intervals where p  x   0 and p  x   0 . e) The symmetry.
[2] 8. Let P = (x , y) be a point on the graph of y  x . Express the distance from P to the point (1 , 0 ) as a
function of x.
[3] 9. Alan is building a garden shaped like a rectangle with a semicircle attached to one side. If he has 40
feet of fencing to go around it, what dimensions will give him maximum area in the garden?
[3] 10. For the function g  x   3 
2
x 1
, find the difference quotient
answer.
[3] 11. If
f  x   x  3 and g  x   x  2
a) Find  f
 g  x  ,
b) Domain of
:
f  x  g  x.
g  x  h  g  x
h
and simplify your
2
[4] 12. Graph the function whose equation is f  x   3 
2
,
x 1
by starting with its basic function and show
and explain each transformation you use to obtain your final graph.
ANSWERS
f  x
f 
5 x
1. f  x   5  x and g  x   x  3 find    x  
, domain of f(x) is: 5  x  0  x  5, domain of

g  x
x3
g
g(x) is: x  3  0

x  3 , therefore the domain of x  3  0

x  3 is x  5 and x  3 , which
gives the final domain as  3,5 or 3  x  5 .
2.
f  x   2 x 2  8 x  4 and g  x   3x 2  12 x  5 , find the solution set for:
f  x  g  x
2 x 2  8x  4  3x 2  12 x  5

 5x  1 x  1  0
5x 2  4 x  1  0

critical values are: x  0.2,1
3.
a
f  x 
a
x
2
DQ
f  x  h  f  x
4. From y  f  x 
h


y
 x  h 2

a
x2
=
h
1
f  2x  ,
2
ax 2  a  x  h 
 x  h
2
2
2
x h
=
ax 2  ax 2  2ahx  ah2
 x  h
2
2
x h
is
A vertical compression and flip by a factor
of  1 2 , horizontal compression and flip by a factor of  1 2 , this gives new
x
4
2
coordinates 0
2
4
6
y
0
3
0
2
2
0
then 2 right and 3 down.
to
x
2
1
0
-1
-2
-3
y
0
-1.5
0
1
1
0
=
2ahx  ah2
 x  h 2  x 2  h
=
2ax  ah
 x  h 2  x 2
3
y
then to
x
4
3
2
1
0
-1
1
f  2  x  2    3
2 
y
-3
-4.5
-3
2
2
-3
f  x     x  5
5.
x
3
5
6
7
Domain:  ,   ,
2
x  -3
y
 4 open
0
1
4
f  x   3
 3  x  5
y
4
5
7
2
x
5
6
7
range:  , 0 , not a function, does not pass vertical line test
6. a) average rate of change is:
1
6
f  x = x  7
x
3
2
0
5
x5
y
3
3
3
5  3 and 5  2 .
5
1
y y2  y1 f  x2   f  x1  f  2   f 1 6
1
Avg rate of change 




 
x x2  x1
x2  x1
2 1
1
6
1
6
b) Msec   , choose point 1,1 , point-slope form: y  y1  m  x  x1   y  1    x  1 
y
1
7
x
6
6
4
7.
p  x  
x
2
4
1
x-1 x  -2
2
y
0 open
1
p  x  =-2 x +6
x
2
0
3
a) Domain:  ,   , range;
decreasing on  , 2 

0,  
p x  3 x  3
2 x3
y
x
2 closed
6
0 closed
y
3
4
7
x3
0 open
3
6
b) max at ( 0, 6 ), min at ( 3, 0 ), c) Increasing on:  2, 0 
 0,3 d)
p  x   0,
x  Reals, x  0, p  x   0, no solution ,
symmetry because of the restricted domains of this piecewise function, if

 3, 0  ,
e) symmetry: there is no
p  x   2 x  6 ,
was restricted
from x = -2 to x = 2, then we would have symmetry over those x values, but this is not the case.
8. The graph looks like:
The distance formula is; D   x2  x1 2   y2  y1 2 , since the equation of the graph is y  x , we can sub
this in for y : D   x  12   x  0   x 2  2 x  1  x  x 2  x  1
2
9. The diagram looks like a Norman window, a rectangle surmounted by a semicircle. If the radius of the
circle is denoted by r , then the width is 2r , the circumference of the semicircle is  r , then the length is
given by; L 
40  2r   r

 20  r  r
2
2
. Then the area is given by:
5
L=20-r-r(π/2)
2r
C=2πr
r
  1
1
1

A  rectangle + semicircle  2r  20  r  r    r 2  40r  2r 2   r 2   r 2  2r 2   r 2  40r
2  2
2
2

1 

A   2    r 2  40r this is a quadratic equation and since "a" is negative the parabola opens down and we have a maximum which occurs at the vertex.
2 

We know the r value at the vertex is: r  
b
40
40
40



~ 5.600991535, and L  20 
 ~ 5.6,
1  4
2a
 4 2

2  2   
2 

therefore the dimensions of the rectangle are 11.2 feet by 5.6 feet and the radius of the semicircle is 5.6
feet.
10.
g  x  h  g  x
2
g  x  3 


x 1
h
2  x  1  2  x  h  1
 x  h  1 x  1



h
1
3
2
2 

 3
x  h  1 
x  1 
h
2 x  2  2 x  2h  2 1
2h
 
 x  h  1 x  1 h  x  h  1 x  1 h
2
 x  h  1 x  1
11. a)
b)
f  x  g  x  x  3  x  2
Domain: domain of f  x   g  x  , domain of f  x  : x  3 0
domain of g  x  : x  2  0

x 2
 x  3

 x  3  x  2



 2,  

12.
I Basic: y 
x
1
y
2
2
1
1
2
1
1
x
y
x
1
2
1
y
1
2
1
2
2
2
asymptotes: x  0, y  0
2
vertical stretch & flip by factor of  2.
x
x
y
x
y
1
1
4
4
2
2
1
2
1
2
2
1
2
1
asymptotes; x  0, y  0
2
 3 1 left, 3 up
x 1
x y
x
y
y
 1.5 7  0.5
1
2 5
0
1
3 4
1
2
asymptotes: x  1, y  3
6