M 115 PRACTICE TEST 2
f 
f ( x)
[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 ) = 2 x2 + 8 x + 4 and g ( x ) = −3x2 + 12 x + 5 , find the solution set for: f ( x ) > g ( x ) . .
[3] 3.* For the function
[3] 4. Given the graph of
f ( x) =
a
x
2
y = f ( x)
f x+h − f x
, find and simplify the difference quotient ( ) ( ) .
h
shown below, show and explain all transformations you use to obtain the
1
graph of y = − f  −2 ( x − 2 ) − 3
2
2
 − ( x + 5) x < −3

−3 ≤ x ≤ 5 .
R ( x) =  x − 7

x≥5
−
3

[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 .
Find the following:
x>3
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
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 +
answer.
[3] 11*. If
a) Find ( f
2
x −1
, find the difference quotient
f ( x ) = x + 3 and g ( x ) = x − 2
+ g )( x ) ,
b) Domain of
g ( x + h) − g ( x)
h
and simplify your
:
f ( x) + g ( x).
Practice Test 2 pg 1
2
, by starting with its basic function and
x −1
[4] 12.* Graph the function whose equation is f ( x ) = 3 −
show
and explain each transformation you use to obtain your fin
final graph.
ANSWERS
f ( x)
 f 
5− x
=
(x) =
g ( x)
x−3
g
1. f ( x ) = 5 − x and g ( x ) = x − 3 find 
, domain of f(x) is: 5 − x ≥ 0
⇒
x ≥ 3 , therefore the domain of x − 3 ≥ 0
g(x) is: x − 3 ≥ 0
gives the final domain as ( 3,5] or 3 < x ≤ 5 .
2.
⇒
x≥3
is
⇒
x ≤ 5,
x ≤ 5 and x ≥ 3 ,
domain of
which
f ( x ) = 2 x 2 + 8 x + 4 and g ( x ) = −3 x 2 + 12 x + 5 , find the solution set for:
f ( x) > g ( x)
2 x 2 + 8 x + 4 > −3 x 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
DQ
2
f ( x + h) − f ( x)
h
4. From y = f ( x )
⇒
=
y=−
( x + h)
2
−
a
x2
=
h
1
f ( −2 x ) ,
2
ax 2 − a ( x + h )
(x + h)
2
2
2
⋅x ⋅h
=
ax 2 − ax 2 − 2ahx − ah 2
(x + h)
2
2
⋅ x ⋅h
=
−2ahx − ah 2
( x + h )2 ⋅ x 2 ⋅ h
=
−2ax − ah
( x + h )2 ⋅ x 2
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 coordinates
to
x
y
x
y
−4
−2
0
3
2
1
0
-1.5
0
0
0
0
2
4
−2
−2
-1
-2
1
1
6
0
-3
0
then 2 right and 3 down.
Practice Test 2 pg 2
y=−
x
then to
1
f  −2 ( x − 2 )  − 3
2 
y
4
3
2
1
-3
-4.5
-3
−2
0
-1
−2
-3
f ( x ) = − ( x + 5)
x
5.
2
f (x)= x − 7
x < -3
y
x
x
−4
−5
6
−3
−3
7
−3
−3
−5
− 4 open
0
−6
−1
0
−7
−7
−4
5
−2
Domain: ( −∞, ∞ ) ,
x≥5
y
5
range: ( −∞, 0 , not a function, does not pass vertical line test
6. a) average rate of change is:
1
6
−3
−2
f ( x ) = −3
− 3 ≤ x ≤ −5
y
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
form: y − y1 = m ( x − x1 ) ⇒ y − 1 = − ( x − 1) ⇒
b) M sec = − , choose point (1,1) , point-slope form
y=−
1
7
x+
6
6
Practice Test 2 pg 3
7.
p ( x) = −
x
1
x-1
2
y
p ( x ) =-2 x +6
x < -2
−2
0 open
−4
1
x
y
−2
0
3
p(x) = 3 x − 3
−2≤ x≤3
x
2 closed
6
0 closed
x>3
y
3
0 open
4
3
7
6
a) Domain: ( −∞, ∞ ) , range; 0, ∞ ) b) max at ( 0, 6 ), min at ( 3, 0 ), c) Increasing on: ( −2, 0 ) ∪ ( 3, ∞ ) ,
decreasing on ( −∞, −2 ) ∪ ( 0, 3) d) p ( x ) ≥ 0, x ∈ Reals; p ( x ) < 0, no solution , e) symmetry: there is no
symmetry because of the restricted domains of this piecewise function, if 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
2
this in for y : D = ( x − 1)2 + ( x − 0 ) = x 2 − 2 x + 1 + x = x 2 − x + 1
9. The diagram looks like a Norman window, a recta
rectangle
ngle surmounted by a semicircle. If the radius of the
circle is denoted by r , then the width is 22r , the circumference of the semicircle is π r , then the length is
given by; L =
40 − 2r − π r
π
= 20 − r − r
2
2
. Then the area is given by:
Practice Test 2 pg 4
L=20-r-r(π/2)
2r
C=2πr
r
π  1
1
1

A = rectangle + semicircle = 2r  20 − r − r  + π r 2 = 40r − 2r 2 − π r 2 + π r 2 = −2r 2 − π r 2 = 40r
2  2
2
2

1 

A =  −2 − π  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
π
=−
=
~ 5.600991535, and L = 20 −
− ~ 5.6,
1  4+π
π +4 2
2a

2  −2 − π 
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
−2 h
⋅ =
( 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
2
y
2
1
1
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 right, 3 up
x −1
x y
x
y
y=
0.5 7
1.5
0 5
2
−1 4
3
−1
1
2
asymptotes: x = 1, y = 3
Practice Test 2 pg 5
Practice Test 2 pg 6