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Solve the equations.
#1
9e x  4  247
#2
7  26 x13  8  216
#3
1  ln x  0
#4
log 2  x   log 2  x  1  3 .
#5
2 x3  x2 13x  6
#6
2 x2  2
2
x 2  3x  4
Find the inverses of the following one-to-one functions.
f  x   53 x  7
#7
#8
#9
L  x   log3  x  2 
Write the general term for the sequences below.
#11
an  8,  1, 6, 13, 20,
A  x   2e5 x
#10
q  x 
#12
bn  8, 32,  128, 512,
2
x
Evaluate the following.
50
#13
  2k  1
#14
k 1
Suppose T :
3

k 1
#15
k 1
2
such that if x 
3
, then T : x
  1 n 
15   

n 1 
  3  

2 
4
 5  3 12 
A

Ax where

.
2 4
 1
 12 
 
#16 Use T to map x   2  .
 4 
 
Graph the following functions.
2x2  7x  3
x 1
#17
P(x) = 3x4 + 5x3 – 64x2 – 74x – 20
#18
r( x) 
#19
y( x)  10 x  1
#20
g ( x)  ln x  2
SOLUTIONS
#1
#2
9e  4  247
x
9e x  243
7  26 x 13  8  216
e x  27
7  26 x 13  224
ln e x  ln 27
26 x 13  32
26 x 13  25
x  ln 27
6 x  13  5
x  ln 33
x  3ln 3
6 x  18
x  3 1.1
x3
x  3.3
#3
1  ln x  0
#4
1  ln x
log 2  x   log 2  x  1  3
 x 
log 2 
3
 x 1 
e x
1
ex
x
x 1
8  x  1  x
23 
8x  8  x
7x  8
x
#5
8
7
#6
2 x3  x 2  13x  6
2 x3  x 2  13x  6  0
 2 x  1 x  2  x  3  0
x   12 , x  2, x  3
2x2  2
2
x 2  3x  4
2 x 2  2  2  x 2  3x  4 
2 x2  2  2 x2  6 x  8
2  6x  8
 6  6x
1  x
#7
#8
y  2e 5 x
x  2e 5 y
x
2
y  53 x  7
ln  2x   ln  e5 y 
x  53 y  7
ln  2x   5 y
x  7  53 y
5
3
 e5 y
 x  7  y
f 1  x   53  x  7 
1
5
ln  2x   y
ln  2x    y


1
5
A1  x   ln
#9
 
5 x
2
#10
y  log 3  x  2 
y
x  log 3  y  2 
x
y  2  3x
2
x
2
y
xy  2
y  3x  2
y
L1  x   3x  2
2
x
q 1  x  
2
x
#11 Note that an is arithmetic with a common difference equal to 7. Use the formula an  ak  d  n  k 
as below.
an  a1  d  n  1
an  8  7  n  1
an  8  7n  7
an  7n  15
#12 Note that bn is geometric with a common ratio equal to 4. Use the formula an  a1r n1 as below.
bn  b1r n 1
bn  8  4 
n 1
50
#13
2 
4
  2k  1
#14
k 1
k 1
k 1
n
 a1  an 
2
50
S50   a1  a50 
2
50
S50  1  99 
2
S50  25 1  99 
Sn 
S50  25 100 
S50  2,500

Sn 
S4 
S4 
S4 
a1 1  r n 
1 r
a1 1  r 4 
1 r
11  24 
1 2
11  16 
1
S 4  1 15 

50
  2k  1  2,500
k 1
  2   15
4
k 1
k 1
  1 n 
15   

n 1 
  3  

#15
a1
1 r
5
S 
1
1
3
5
S 
3 1

3 3
5
S 
2
3
3
S  5 
2
15
S 
2

S 
  1  n  15
15    

n 1 
  3   2

#16
 12 
 12 
1
   5 3 2 
T  2   
 2
2 4   
 4   1
 
 4 
 5   3   1 
 12    2    4  2 
 1   2   4 
 60   6    2 

 

 12   4   16 
 12 
   64 
T  2   

 4   32 
 
#17 Graph P(x) = 3x4 + 5x3 – 64x2 – 74x – 20
Polynomial functions have a domain that includes all real numbers. Determine that there is one positive
real root and three (or one) negative real roots using Descartes's Rule of Signs. Determine also that the
possible rational roots are ±1, ±2, ±4, ±5, ±10, ±20, ±⅓, ±⅔, ±4/3, ±5/3, ±10/3, ±20/3.
Use synthetic division to find the rational roots. Use numbers from the list above. Divide until getting
zero as a remainder.
-5 3 5
-64 -74 -20
-15 50 70 20
3 -10 -14 -4
0
Synthetic division with –5 as the divisor reveals that –5 is a root and that 3x3 – 10x2 – 14x – 4 is a cubic
factor of the polynomial. Divide into 3x3 – 10x2 – 14x – 4 next in order to reduce the polynomial to a
quadratic factor.
– 2/3 3 -10 -14 -4
-2 8
4
3 -12 -6
0
– ⅔ is a root, and 3x2 – 12x – 6 is a quadratic factor of the polynomial. To find the remaining two roots,
set the quadratic equal to zero and solve.
3x 2  12 x  6  0
There is a greatest common factor so reduce the equation.
3x 2 12 x 6 0

 
3
3
3 3
x2  4x  2  0
Use the quadratic equation to solve.
x
  4  
 4 2  41 2 
21
4  16  8
2
4  24
x
2
4  46
x
2
42 6
x
2
4 2 6
x 
2
2
x 2 6
x
Now that all the roots are found, we see that they are all real & unique roots, meaning there are four xintercepts. One on the positive side of the x-axis at x  2  6 , and three on the negative side of the xaxis at x = –5, x = – ⅔, and x  2  6 .
The y-intercept is always at the constant, so this polynomial has a y-intercept at (0, –20).
The right end is going up because the leading coefficient is positive. The left end is also going up because
the degree is even so the polynomial has same end behavior
2 
 5, 0   
2
3
, 0
6, 0

2 
6, 0

 0, 20
#18 r ( x ) 
2x2  7x  3
x 1
First, determine that the domain is all real numbers except one. One is the only restriction.
2 x 2  7 x  3 2 x  1x  3
Second, factor the numerator and see if the function reduces: r( x ) 
.

x 1
x 1
Since the function does not reduce, the restrictions represent vertical asymptotes. So, the function has a
vertical asymptote at x = 1.
Third, determine that there is no horizontal asymptote because the degree of the numerator is larger than
the degree of the denominator. Recall the rules for horizontal asymptotes. The rational function on the
final exam may have a horizontal asymptote. This example does not.
Fourth, determine that this rational function has a slant asymptote since the degree of the numerator is
exactly one larger than the degree of the denominator and the denominator does not divide evenly into the
numerator. Divide to find the slant asymptote. The asymptote will be represented by the equation y =
quotient without remainder. Since the quotient is 2x + 9, the slant asymptote is y = 2x + 9.
2x  9
x  1 2x  7x  3
2
2
– 2 x  2 x 
9x  3
9 x  9 
12
–
Fifth, determine the y-intercept by substituting zero for x.
2x2  7x  3
r( x) 
x 1
2
20  70  3
r ( 0) 
0 1
3
r ( 0) 
1
r ( 0 )  3
(0,3)
Sixth, determine the x-intercepts by setting the numerator equal to zero.
2x2 + 7x + 3 = 0
(2x + 1)(x + 3) = 0
2x + 1 = 0
x+3=0
x = –½
x = –3
x-intercepts: (–½,0) & (–3,0)
Seventh, generate a table of ordered pairs then graph the function.
x
r(x)
–10 –12.09
–5 –3
–3 0
–1 1
–½ 0
0
–3
½
–14
⅞
–85.25
1⅛ 107.25
1½ 36
3
21
10 30.33
The purple curves represent the function r(x).
Note that there are two curves, one on either side
of the vertical asymptote. Rational functions with
vertical asymptotes are discontinuous.
This dotted blue oblique line represents the slant
asymptote. SA: y = 2x + 9
This solid purple vertical line represents the
vertical asymptote. VA: x = 1
#19 y( x)  10 x  1
Exponential functions of the form y = bx where b > 0, b ≠ 1 have a horizontal asymptote of y = 0 unless
there is a shift up/down. This function has a shift down one without any reflection, so the horizontal
asymptote is y = –1.
Evaluate y(0) to find the y-intercept:
y (0)  10 0  1
y ( 0)  1  1
y ( 0)  0
In this function the origin, (0,0), is the y-intercept and the x-intercept. If the x-intercept is a distinct point
from the y-intercept, setting the function equal to zero and solving finds the x-intercept as shown below.
10 x  1  0
10 x  1
log 10 x  log 1
x0
Generate a table and sketch the exponential curve.
x
-2
-1
0
1
2
y(x)
-.99
-.9
0
9
99
The purple curve represents the function y(x).
The dotted blue horizontal line represents the horizontal asymptote. HA: y = –1
#20 g ( x)  ln x  2
Logarithmic functions of the form y = logb(x + c) where b > 1, have a domain restricted by the equality x
+ c > 0. This function, therefore, has a domain given by x > –2.
These functions also have a vertical asymptote represented by the linear equation x + c = 0. The function
g(x) = ln(x + 2), therefore, has a vertical asymptote at x = –2.
Evaluate g(0) to find the y-intercept:
g (0)  ln( 0  2)
g (0)  ln( 2)
g (0)  0.69
The function intersects the y-axis at the point (0, ln2).
Setting the function equal to zero and solving finds the x-intercept:
ln( x  2)  0
x  2  e0
x 2 1
x  1
The function intersects the x-axis at the point (–1, 0).
Generate a table and sketch the logarithmic curve. Remember the domain and use x-values greater than –
2.
x
g(x)
-1.999 ln(.001) ≈ -6.9
-1.9
ln(.1) ≈ -2.3
-1
0
0
ln(2) ≈ .69
1
ln(3) ≈ 1.1
2
ln(4) ≈ 1.4
The dotted blue vertical
line represents the
vertical asymptote.
VA: x = –2
The purple curve represents the function g(x).