Calculus 1
Review- Units P & 1 Test
Name:______________________________
Date:_______________________________
SHOW ALL WORK!
(1) Which of the following equations define functions? Explain your reasoning.
(a) y = x + 6
Function- Passes VLT
(b) y2 = x + 1
Not a Function- Fails VLT
(c) y3 = x + 4
Function- Passes VLT
(d) y = x – 2
Function- Passes VLT
(e) y3 = x – 3
Function- Passes VLT
(f) y2 = x – 5
Not a Function- Fails VLT
(2) Which of the following functions are one-to-one? Explain your reasoning.
(a) f(x) = x5
One-to-one- Passes both VLT and HLT
(b) g(x) = 2x + 7 One-to-one- Passes both VLT and HLT
(c) h(x) = x2 – 4 NOT One-to-one- FAILS HLT
(3) Write the linear equation in standard form Ax + By = C that passes thru the points (3, 5) and (4, -3).
35
 8
43
y  mx  b
m
y  8 x  29
 8 x  y  29
I choose (3,5)
(5)  ( 8)(3)  b
5  24  b
b  29
(4) Write the linear equation in slope-intercept form y = mx + b that passes through the point (1, 3) and is
parallel to the line 2x + 2y = 5.
2x + 2y = 5
2y  -2x  5
y  -x  2.5
 m  -1
Parallel lines have = slopes.
 y  x  4
y  mx  b
(3)  (1)(1)  b
3  1  b
b4
1
(5) Write the linear equation in slope-intercept form y = mx + b that passes through the point (1, 3) and is
perpendicular to the line 3x – 3y = 4.
3x - 3y = 4
y  mx  b
- 3y  -3x  4
(3)  (1)(1)  b
yx
 y  x  4
3  1  b
4
3
b4
m 1
 the NEW m  1
(6) Write the linear equation in standard form Ax + By = C that is a horizontal line and passes through the
point (-9, 2).
y=2
__________________________________________________________________________________________
(7) Find the point(s) of intersection between: y  2( x  3) 2  4 & 3 y  4 x
Algebraic Approach: Substitution
y  2( x  3) 2  4 & 3 y  4 x


3 2( x  3) 2  4  4 x
6( x  3) 2  12  4 x
6( x 2  6 x  9)  12  4 x
Now sub each x-value into one of the two equations to
find the corresponding y-value
3y  4x
If x=11/3, then y=44/9
If x=3, then y=4
Therefore, (11/3, 44/9) and (3,4) are our solutions
6 x 2  36 x  54  12  4 x
6 x 2  40 x  66  0
3 x 2  20 x  33  0
(3 x  11)( x  3)  0
x  11 / 3 x  3
(8) Given f(x) = 2x2 – x + 3 find f(k – 2)
f(k - 2)  2(k - 2) 2  (k  2)  3  2(k 2  4k  4)  k  2  3  2k 2  8k  8  k  2  3  2k 2  9k  13
(9) Given f(x) = 4x – 7 and g(x) = 2x – x2, evaluate f(2) + g(-1)
f(2) = 4(2) - 7 = 1
g(-1) = 2(-1) - (-1) 2  3
1  3  2
2
For the function f(x) = -6x + 5, find and simplify:
(10)
(a)
f ( x  h)  f ( x )
, h0
h
f ( x  h )   6 ( x  h )  5  6 x  6 h  5
f ( x )  6 x  5
f ( x  h)  f ( x )

h
 6 x  6h  5  (6 x  5)

h
 6 x  6h  5  6 x  5

h
 6h
 6
h

(11) Find the domain of each of the following:
2x  3
(a) f ( x ) 
2x
(b)
f ( x )  f (a )
, x a  0
x a
 6 x  5  (6a  5)

xa
 6 x  5  6a  5

xa
 6 x  6a

xa
 6( x  a )
 6
xa
(b) g( x )  x  4
x  2
x4
or
or
 ,2   2,  
4, )
(12) Algebraically determine the symmetry with respect to the y-axis, x-axis, and origin, if any
exists, for each of the given equations:
(a) 2x – 4y = 7
(b) 9x2 – 4y2 = 36
RULES: If you replace x with –x and simplify back to the original equation, then function has y-axis symmetry
If you replace x with –y and simplify back to the original equation, then function has x-axis symmetry
If you replace x&y w/ –x&–y and simplify back to original eq., then function has x-axis symmetry
(a) NONE
(b) ALL
(13) Algebraically determine if each of the given functions are odd, even, or neither:
(a) f(x) = x2 – 6x
(b) f(x) = x6 + 7
RULES:
If f(-x)=f(x), then function is EVEN.
If f(-x)=-f(x), then function is ODD.
a) Neither
b) EVEN
3
(14) Perform the four basic operations with the functions f(x) = 4x and g(x) = x2 + 2
(f + g)(x) = x 2  4 x  2
(f – g)(x) =  x 2  4 x  2
(f  g)(x) = 4 x 3  8 x
f
4x
 ( x )  2
x 2
 g
(15) Given f(x) = x2 – 3 and g( x )  2x  4 , find f  g( x) and g f ( x) and give each domain.
 f  g ( x)   2 x  4   3  2 x  4  3  2 x  7
g  f ( x)  2( x 2  3)  4  2 x 2  10
2
D :  ,  

 
D :  , 5 
5, 

(16) For each of the following functions, f(x), find the inverse, f -1(x):
(a) f(x) = 5x + 2
4
(b) f ( x ) 
x3
Switch x and y then solve for y.
a) y= 1/5 x – 2/5
b) y = 4/x – 3
(17) Given the graph on the right, answer the following questions:
(a)
write the linear equation in slope-intercept form:
y = 2x+4
(b)
write in slope-intercept form the equation of the parallel line
through (2, -1) and graph:
y = 2x -5
(c)
y
x
write in slope-intercept form the equation of the perpendicular
line through the x-intercept and graph:
y = -1/2 x – 1
_______________________________________________________________________________________
4
Graph each function and use interval notation to answer the following questions:
(a) Find the domain
(d) Find the intervals over which the function is decreasing
(b) Find the range
(e) Find the intervals over which the function is constant
(c) Find the intervals over which the function is increasing
(f) State any points of discontinuity
2  x
(19) f ( x )  
2
 x
(18) f ( x)  2 x  3  2
if  1  x  0
if 0  x  1
y
y
2
5
x
0
x
0
-8
-6
-4
-2
-3
0
2
4
6
-2
-1
0
1
2
3
8
-2
-5
a)
b)
c)
d)
e)
f)
a)
b)
c)
d)
e)
f)
 , 
 ,2
 ,3
 3, 
[1,0)  [0,1]
[0,1]  (2,3]
(0,1)
(-1,0)
none
x=0
none
none
 1  x
(20) f ( x )  
 5x
if x  2
 x  3 if x  0
(21) f ( x )  
3 x  1 if x  0
if x  2
y
y
5
5
x
0
-8
-6
-4
-2
0
2
4
6
-8
-6
-4
-5
a)
b)
c)
d)
e)
f)
(,2]  (2, )
(, )
none
(, )
none
x=2
x
0
8
-2
0
2
4
6
8
-5
a)
b)
c)
d)
e)
f)
(,0)  [0, )
(, )
(, )
none
none
x=0
5
1 if x  2


(22) f ( x )  2  x if  2  x  2
  2x if x  2

 2x

(23) f ( x )  
3
3x  2

if x  2
if x  2
if x  2
y
y
5
5
x
0
x
0
-8
-6
-4
-2
0
2
4
6
-8
-6
-4
-2
0
2
4
6
8
8
-5
-5
(,2]  (2,2]  (2, )
(,4)  (0,4]
(-2,2]
(2, )
(,2]
x=2, -2
(24) Graph the following piecewise function:

x2
if  2  x  0

if 0  x  2
 4  2x
f ( x)    2
if
x2
 1
  x  1 if 2  x  4
 2
a)
b)
c)
d)
e)
f)
a)
b)
c)
d)
e)
f)
(,2)  [2]  (2, )
(0, )
(2, )
(,2)
none
x=2
y
4
2
x
0
-4
-3
-2
-1
0
1
2
3
4
-2
-4
(25) Based on your graph of f(x) from question (7), find the value of each of the following limits:
(a) lim  f ( x ) = 4
(e) lim f ( x) = 0
x  2
x 2
(b) lim f ( x) = 2
(f) lim f ( x ) = 4
(c) lim f ( x ) = -0.5
(g) lim f ( x ) = 0
(d) lim f ( x ) = -1
(h) lim f ( x) = DNE
x 1
x 3
x 4
x 0
x 0
x 0
6
(26) Based on the given graph of f(x), find the value of each of the following:
(a) lim f ( x ) = 4
y
x 4
(b) lim f ( x ) = 1
x 4
(c) lim f ( x) = DNE
x 4
(d) lim f ( x) = 2
x 1
(e) lim f ( x) = 3
x 3
x
(f) f(3) = 5
(g) lim f ( x) = -1
x  
(h) lim f ( x) = 
x  
Algebraically determine the value each of the following limits or explain why it does not exist:
(27)
(29)
x3

x  3 x 2  9
x3
lim

x  3 ( x  3)( x  3)
1
1
lim

 1 / 6
x  3 x  3
33
(28)
 2x  4
 2(2)  4
8


3
2
3
2
0
x  2x
(2)  2(2)
SO ...WE NEED TO SIMPLIFY
(30)
lim
lim
x 2
lim
x 2
lim
x 2
lim
x 
2x 2
 2 / 3
x  3x 2
m=n
lim
x  
3x  1
 3 / 4
3  4x
m=n
 2( x  2)

x 2 ( x  2)
2
2

 1 / 2
2
x
(2) 2
7
(31)
x3  4x
2 3  4(2) 0


x 2 x 3  2 x 2
2 3  2(2) 2 0
SO ...WE NEED TO SIMPLIFY
lim
(32)
lim
x 0
x
x  6x  9  9
lim

x 0
x
x 2  6x
lim

x 0
x
lim x  6  6
2
x( x  2)( x  2)

x 2
x 2 ( x  2)
x2 22
lim

2
x 2
x
2
lim
(33)
(35)
2 x 2  3x
lim
 2/3
x   3 x 2  48
mn
lim
x  
3x  4
0
x2 1
 x  3 2  9 
x 0
(34)
lim
x  
x2  2
 DNE
2x
mn
(36)
mn
x2 1
(1) 2  1
0


2
2
x  1 2 x  5 x  3
2(1)  5(1)  3 0
SO ...WE NEED TO SIMPLIFY
lim
( x  1)( x  1)

x  1 ( 2 x  3)( x  1)
x 1
11
2
lim


 2
x  1 2 x  3
2(1)  3
1
lim
(37)
(39)
x 2  64
 DNE
x 
x 8
mn
(38)
x3  x 2  2x
lim

x  1
x 2  2x
(1) 3  (1) 2  2(1) 0

 0
3
(1) 2  2(1)
(40)
lim
lim
x  
x4  7
 DNE
8  x3
mn
lim
x 2
3x  18 3(2)  18  24 3



x 2  36 (2) 2  36  32 4
8