MATH1051: Exam 4
Vishal Saraswat
Here is the key to the fourth exam. This exam covered chapter 5 from the book. Be sure you understand how to do
each problem since you will see similar problems on the final exam.
1. (10 points) Find the effective rate of interest on an account whose annual rate is 8% compounded quarterly.
Round your answer to the nearest tenth of a percent.
Solution: Effective Rate of Interest re is the equivalent annual simple rate of interest that would
yield the same amount as compounding after 1 year.
Simple Interest: SI = P re t;
Compound Interest: CI = P
h
1+
r tn
n
i
−1 ;
The time is one year so we replace t = 1.
The rate of interest is given as 8% so we replace r = 0.08.
The interest is compounded is quarterly so n = 4.
Now, SI = CI if and only if,
P re t = P
r tn
1+
−1
n
so that
1 r tn
re =
1+
−1
t
n
"
#
1·4
1
0.08
=
1+
−1
1
4
= (1 + 0.02)4 − 1
= (1.02)4 − 1
= 0.08243216 .
So, the effective rate of interest is 8.243216%, which rounds to 8.2%.
Ans: 8.2%
Vishal Saraswat
MATH1051: Precalculus I, Fall 2009
Page 1 of 8
MATH1051: Exam 4
Vishal Saraswat
x+1
x
and g (x) =
:
x+1
x
(a) (10 points) Find the composition (f ◦ g) (x) and simplify.
2. Given the functions f (x) =
Solution:
Use g(x) as the input for f (x) and then simplify:
y=
(f ◦ g) (x) = f (g (x))
x+1
=f
x
x+1
x
= x+1
+1
x
=
=
x+1
x
(x+1)+x
x
x+1
x
2x+1
x
4
(−2, 2)
(−3, 3/2)
y=
−x
−4
−3
−2
3
(1/2, 3)
2
(1, 2)
1
(−2, 1/2)
x+1
2x+1
−1
(2, 3/2) y =
y=
(1, 1/2)
O
1
x+1
x
2
3
x
x+1
x+1
y = 2x+1
x
4
(−1/2, −1)−1
x+1
=
.
2x + 1
Ans:
y
x
x+1
−2
x+1
2x+1
−3
−4
y=
−y
x+1
x
(b) (5 points) What is the domain of (f ◦ g) (x)?
Solution:
Domain of g(x):
{x|x 6= 0} .
Domain of
x+1
:
2x+1
{x|2x + 1 6= 0} = {x|x 6= −1/2} .
Combining these we have the domain of (f ◦ g)(x):
{x|x 6= 0 and x 6= −0.5} .
Ans: {x|x 6= 0 and x 6= −0.5}
Vishal Saraswat
MATH1051: Precalculus I, Fall 2009
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MATH1051: Exam 4
Vishal Saraswat
1−x
:
1+x
(a) (10 points) Find the inverse f −1 (x) .
3. Given the function f (x) =
Solution:
Set y = f (x) and then solve for x in terms of y.
y
1−x
y=
1+x
or (1 + x)y = 1 − x
or y + yx = 1 − x
or x + yx = 1 − y
or x(1 + y) = 1 − y
1−y
.
or x =
1+y
3
2
1
−x
So,
f −1 (x) =
4
1−x
= f (x) .
1+x
Ans:
−4
−3
−2
−1
(0, 1)
O
−1
(−3, −2)
1−x
1+x
(1, 0)
1
2
3
4
x
y=
1−x
1+x
−2
(−2, −3)
−3
−4
y=
1−x
1+x
−y
(b) (5 points) Find the range of f (x).
The range of f (x) is the domain of f −1 (x). Here f −1 (x) = f (x) so that the range
of f (x) is same as its domain:
Solution:
{x|1 + x 6= 0} = {x|x 6= −1} .
Ans: {x|x 6= −1}
Vishal Saraswat
MATH1051: Precalculus I, Fall 2009
Page 3 of 8
MATH1051: Exam 4
Vishal Saraswat
4. Graph:
(a) (10 points) 2−x ;
(b) (10 points) log2 x.
Solution:
y
(−2, 4)
4
3
y = log2 x
(−1, 2)
2
(4, 2)
1 (0, 1)
(2, 1)
(1, 1/2)
−x
−4
−3
−2
−1
O
−1
1
(1, 0)
x
2
3
4
y = 2−x
(1/2, −1)
−2
−3
−4
−y
Vishal Saraswat
MATH1051: Precalculus I, Fall 2009
Page 4 of 8
MATH1051: Exam 4
Vishal Saraswat
5. (10 points) Calculate
Solution:
3
log √
3
1
.
27
Use the change of base formula:
• loga M =
logb M
.
logb a
Then,
3
log √
3
1
27
log3
1
27
√
log3 3 3
log3 3−3
=
log3 31/3
−3
=
1/3
= −9 .
=
since log3 x is the inverse of 3x
Ans: −9
Vishal Saraswat
MATH1051: Precalculus I, Fall 2009
Page 5 of 8
MATH1051: Exam 4
Vishal Saraswat
"
6. (10 points) Simplify log
Solution:
3x (x + 1)
2
2 (x − 1)
#
.
Use the these properties of logs:
• Log of a product: loga M N = loga M + loga N ;
• Log of a quotient: loga
M
N
= loga M − loga N ;
• Log of a power: loga M r = r loga M .
Then,
3x (x + 1)
2
=
log
[3x
(x
+
1)]
−
log
2
(x
−
1)
log
2 (x − 1)2
= log 3 + log x + log (x + 1) − log 2 + log (x − 1)2
= log 3 + log x + log (x + 1) − [log 2 + 2 log (x − 1)]
= log 3 + log x + log (x + 1) − log 2 − 2 log (x − 1)
= log (x + 1) + log x − 2 log (x − 1) + log 3 − log 2 .
Ans: log (x + 1) + log x − 2 log (x − 1) + log 3 − log 2
Vishal Saraswat
MATH1051: Precalculus I, Fall 2009
Page 6 of 8
MATH1051: Exam 4
Vishal Saraswat
2
2
7. (10 points) Solve: 27 (9x ) = 3−x .
Solution:
Write the expressions so that they have the same base:
2
27 (9x )2 = 3−x
x 2
2
or, 33 32
= 3−x
2x
2
or, 33 32
= 3−x
or, 33 34x = 3−x
2
2
or, 34x+3 = 3−x .
Since the exponential function is one-to-one we get 4x + 3 = −x2 and we solve for x from this
equation.
4x + 3 = −x2
or, x2 + 4x + 3 = 0
or, (x + 3)(x + 1) = 0
or, x = −3 or x = −1 .
Ans: x = −3 or x = −1
Vishal Saraswat
MATH1051: Precalculus I, Fall 2009
Page 7 of 8
MATH1051: Exam 4
Vishal Saraswat
8. (10 points) Solve: log2 (x − 3) = 1 + log4 (x).
Solution:
Use the properties of logs to combine all the log terms into a single term:
or,
or,
or,
or,
log2 (x − 3) = 1 + log4 x
log2 (x − 3) − log4 x = 1
log2 x
=1
log2 (x − 3) −
log2 4
log2 x
log2 (x − 3) −
=1
2
2 log2 (x − 3) − log2 x = 2
or, log2 (x − 3)2 − log2 x = 2
(x − 3)2
=2
x
(x − 3)2
or,
= 22
x
or, (x − 3)2 = 4x
or, log2
or, x2 − 6x + 9 = 4x
or, x2 − 10x + 9 = 0
or, (x − 1)(x − 9) = 0
or, x = 1 or x = 9 .
Since x = 1 is not in the domain of the original equation (log(x − 3) is defined only for x > 3),
the only solution is x = 9.
Ans: x = 9
Vishal Saraswat
MATH1051: Precalculus I, Fall 2009
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