College Algebra
Chapter 5 Review
Created by: Lauren Atkinson
Math Coordinator, Mary Stangler Center for Academic Success
Note:
This review is composed of questions from the
chapter review at the end of chapter 5. This
review is meant to highlight basic concepts from
chapter 5. It does not cover all concepts
presented by your instructor. Refer back to your
notes, unit objectives, handouts, etc. to further
prepare for your exam.
Use the graph to evaluate each expression.
a) (𝑓 − 𝑔)(2)
b) (𝑓𝑔)(0)
𝑦 = 𝑓(𝑥)
𝑦 = 𝑔(𝑥)
a) 𝑓 − 𝑔 2 = 𝑓 2 − 𝑔 2 = 4 − 0 = 4
b) 𝑓𝑔 0 = 𝑓 0 ∙ 𝑔 0 = 0 ∙ −2 = 0
Use 𝑓 𝑥 = 𝑥 2 + 3𝑥 and 𝑔 𝑥 = 𝑥 2 − 1 to find
each expression. Identify its domain.
𝑓 + 𝑔 𝑥 = 𝑓 𝑥 + 𝑔 𝑥 = 𝑥 2 + 3𝑥 + 𝑥 2 − 1
𝑓+𝑔 𝑥
= 2𝑥 2 + 3𝑥 − 1 domain= all real numbers
a) (𝑓 − 𝑔)(𝑥)
b) 𝑓𝑔 𝑥
c)
𝑓
( )(𝑥)
𝑔
𝑓 − 𝑔 𝑥 = 𝑓 𝑥 − 𝑔 𝑥 = 𝑥 2 + 3𝑥 − (𝑥 2 − 1)
= 𝑥 2 + 3𝑥 − 𝑥 2 + 1 = 3𝑥 + 1
domain= all real numbers
𝑓𝑔 𝑥 = 𝑓 𝑥 ∙ 𝑔 𝑥 = (𝑥 2 + 3𝑥)(𝑥 2 − 1)
= 𝑥 4 − 𝑥 2 + 3𝑥 3 − 3𝑥 = 𝑥 4 + 3𝑥 3 − 𝑥 2 − 3𝑥
Domain= all real numbers
𝑓
𝑔
𝑥 =
𝑓(𝑥)
𝑔(𝑥)
=
𝑥 2 +3𝑥
𝑥 2 −1
Domain= set the denominator equal to 0 and solve
𝑥 2 − 1 = 𝑥 + 1 𝑥 − 1 = 0 so 𝑥 = ±1
All real numbers except 𝑥 ≠ ±1
Use the graph to evaluate each expression.
a) (𝑓 ∘ 𝑔)(2)
b) 𝑔 ∘ 𝑓 0
c) 𝑓 −1 (1)
𝑦 = 𝑓(𝑥)
𝑦 = 𝑔(𝑥)
a) 𝑓 ∘ 𝑔 𝑥 = 𝑓 𝑔 𝑥
which in our case means 𝑓 𝑔 2 . 𝑔 2 = −2 so 𝑓 −2 = 0
b) 𝑔 ∘ 𝑓 𝑥 = 𝑔 𝑓 𝑥
which in our case means 𝑔 𝑓 0 . 𝑓 0 = 2 so 𝑔 2 = −2
c) 𝑓 −1 means the inverse of 𝑓. So instead of trying to find the y-coordinate when 𝑥 is
1, we find the inverse, or the x-coordinate when 𝑦 is 1: 𝑓 −1 1 = −1
Use 𝑓 𝑥 = 𝑥 2 + 1 and 𝑔 𝑥 = 𝑥 3 − 𝑥 2 + 2𝑥 +
1 to find each expression.
a) (𝑓 ∘ 𝑔)(𝑥)
b) (𝑔 ∘ 𝑓)(𝑥)
a) 𝑓 ∘ 𝑔 𝑥 = 𝑓(𝑔 𝑥 ) meaning we plug the entire function 𝑔, into 𝑓, whenever we
see an x: 𝑓 ∘ 𝑔 𝑥 = (𝑥 3 − 𝑥 2 + 2𝑥 + 1)2 +1
b) 𝑔 ∘ 𝑓 𝑥 = 𝑔 𝑓 𝑥 meaning we plug the entire function 𝑓, into 𝑔, whenever we
see an x: 𝑔 ∘ 𝑓 𝑥 = (𝑥 2 + 1)3 − 𝑥 2 + 1 2 + 2 𝑥 2 + 1 + 1
Find 𝑓 ∘ 𝑔 𝑥 and identify it’s domain.
a) 𝑓 𝑥 = 𝑥 + 3
𝑔 𝑥 = 1 − 𝑥2
𝑓∘𝑔 𝑥 =𝑓 𝑔 𝑥 =
1 − 𝑥2 + 3 = 4 − 𝑥2
Domain= since we are dealing with a square root, we know that 4 − 𝑥 2 has to be greater
than zero (otherwise we don’t have real solutions)
4 − 𝑥2 ≥ 0
4 ≥ 𝑥 2 ±2 ≥ 𝑥 −2 ≤ 𝑥 ≤ 2 so our domain is 𝑥 −2 ≤ 𝑥 ≤ 2
b) 𝑓 𝑥 =
𝑓∘𝑔 𝑥 =𝑓 𝑔 𝑥
2
𝑥−5
=
2
1
−5
𝑥+1
𝑔 𝑥 =
1
𝑥+1
2𝑥+2
= −4−5𝑥 (there was a bit of algebra involved here)
Domain= we are only concerned with the denominator. −4 − 5𝑥 ≠ 0 so we set it equal to
4
zero and solve: −4 − 5𝑥 = 0 −4 = 5𝑥 − 5 = 𝑥 but we also have to look at the original
function we plugged in (𝑔) and be careful about that denominator as well. 𝑥 + 1 = 0
4
4
𝑥 = −1 . So the domain is going to be all real numbers except 𝑥 = −1, − 5 = 𝑥 𝑥 ≠ −1, − 5
Find 𝑓 and 𝑔 so that ℎ 𝑥 = (𝑔 ∘ 𝑓)(𝑥).
ℎ 𝑥 =
1
(2𝑥+1)2
There are several options for this question, however, the easiest one is as follows:
1
1
𝑔 𝑥 = 𝑥 2 and 𝑓 𝑥 = 2𝑥 + 1 so when we find (𝑔 ∘ 𝑓)(𝑥) is becomes (2𝑥+1)2 .
Determine if 𝑓 is one-to-one.
𝑓 𝑥 = 3𝑥 2 − 2𝑥 + 1
Usually to determine if a function is one-to-one, we graph the function and see if the
function passes the horizontal line test.
This graph doesn’t pass the horizontal
line test, therefore it IS NOT ONE-TO-ONE.
Describe the inverse operations of the given
statement. Then express both the statement
and its inverse symbolically.
Subtract 5 from 𝑥 and take the cube root.
The inverse of the statement would be to cube 𝑥 and then add 5.
Written out that looks like this: 𝑥 3 + 5
The statement and its inverse symbolically:
3
Statement: 𝑥 − 5
Inverse: 𝑥 3 + 5
Determine if f is one-to-one
This graph does pass the horizontal line test, therefore, the function IS ONE-TO-ONE
Find 𝑓 −1 𝑥 .
𝑓 𝑥 =
3𝑥
𝑥+7
3𝑥
We can rewrite this as 𝑦 = 𝑥+7 and then flip our variables, and solve for y:
3𝑦
𝑥 = 𝑦+7
𝑥 𝑦 + 7 = 3𝑦
𝑥𝑦 + 7𝑥 = 3𝑦
7𝑥 = 3𝑦 − 𝑥𝑦
7𝑥 = 𝑦(3 − 𝑥)
7𝑥
=𝑦
3−𝑥
7𝑥
𝑓 −1 𝑥 = 3−𝑥
Use the tables to evaluate the given expression.
𝑥
0
1
2
3
𝑓(𝑥) 4
3
2
1
𝑥
0
1
2
3
𝑔(𝑥) 0
2
3
4
(𝑔−1 ∘ 𝑓 −1 )(1)
To make this easier on ourselves, I am going to find the tables for 𝑓 −1 (𝑥) and 𝑔−1 𝑥 :
To do this, we just flip the values in our tables from above,
𝑥
4
3
2
1
𝑓 −1 (𝑥) 0
1
2
3
𝑥
0
2
3
4
𝑔−1 (𝑥) 0
1
2
3
Then we can find (𝑔−1 ∘ 𝑓 −1 )(1) by finding 𝑔−1 𝑓 −1 1 . 𝑓 −1 1 = 3 and 𝑔−1 3 = 2
so 𝑔−1 ∘ 𝑓 −1 1 = 2
Simplify 𝑒 𝑥 𝑒 −2𝑥 .
Since the bases of our exponents are the same, we can add our exponents:
𝑥 + −2𝑥 = 𝑥 − 2𝑥 = −𝑥
= 𝑒 −𝑥
Find C and a so that 𝑓 𝑥 = 𝐶𝑎 𝑥 satisfies the
given conditions.
𝑓 −1 = 8 and 𝑓 1 = 2
We have to use the conditions above to formulate some equations:
8 = 𝐶𝑎 −1 and 2 = 𝐶𝑎1
𝐶
So we know that 8 = 𝑎 or that 8𝑎 = 𝐶 and we can plug that into our second equation,
2 = (8𝑎)𝑎1
2 = 8𝑎2
1
1
2
=
𝑎
=𝑎
4
2
And then we can plug it back in to solve for C: 8
1
So 𝐶 = 4 and 𝑎 = 2
1
2
=𝐶
4=𝐶
Use the graph of 𝑓 𝑥 = 𝐶𝑎 𝑥 to determine
values for C and a.
We can pull two points off of the graph to help us determine 𝐶 and 𝑎: (0,3) and (1,1)
3 = 𝐶𝑎0 3 = 𝐶
So now we know that 𝑦 = 3𝑎 𝑥 and we can plug in the point (1,1) to figure out what 𝑎 is.
1
1 = 3𝑎1 1 = 3𝑎 3 = 𝑎
1
So 𝐶 = 3 and 𝑎 = 3
Determine the final value of $500 invested at
6.5%, compounded continuously for 8 years.
We need to use the compound interest formula: 𝐴 = 𝑃𝑒 𝑟𝑡
Where 𝑃 = $500, 𝑟 = 6.5% 𝑜𝑟 .065, 𝑡 = 8
𝐴 = 500𝑒 .065∙8 = 500𝑒 .52 = $841.01
Evaluate the expression without a calculator.
a) log 0.001
3
b) log 100 + log 10
a) Without a calculator, we ask ourselves, 10 to what power of x will give us 0.001?
or 10? = 0.001. We think about moving the decimals and notice that in this case we
move the decimal 3 places to the left so 10−3 = 0.001.
= −3
3
b) Again, 10? = 100 and 10? = 10. These are a bit easier than the question before
102
1
3
3
because we know that
= 100 and 10 = 10 therefore, our equation changes
1
1
7
7
3
from log 100 + log 10 to 2 + 3 and we know that 2 + 3 = 3 = 3
Evaluate the logarithm without a calculator.
a)
1
log 5
25
b) log 2 32
1
1
a) Without a calculator, we ask ourselves, 5? = 25 and we figure out that 5−2 = 25 so
1
log 5 25 = −2
b) Again, we ask, 2? = 32 and we figure out that 24 = 32 so log 2 32 = 4
Approximate to the nearest thousandth.
log 2 173
We can’t just plug this into our calculator because our calculators work in base 10. In order
to approximate this, we must use the change of base formula:
log 2 173 =
log 173
log 2
= 7.435
Solve the equation.
1.5𝑥 = 55
log 1.5𝑥 = log 55
𝑥 log 1.5 = log 55
log 55
𝑥 = log 1.5 = 9.883
4𝑒 2𝑥 − 5 = 3
4𝑒 2𝑥 = 8
𝑒 2𝑥 = 2
ln 𝑒 2𝑥 = ln 2
2𝑥 = ln 2
𝑥=
ln 2
2
= 0.3466
Solve the equation.
3 10 −6 = 6
10 −6 = 2
log 10−𝑥 = log 2
−𝑥 log 10 = log 2
−𝑥 = log 2
𝑥 = − log 2 = −.301
5 1.3
𝑥
+ 4 = 104
5 1.3 𝑥 = 100
1.3 𝑥 = 20
log 1.3 𝑥 = log 20
𝑥 log 1.3 = log 20
log 20
𝑥 = log 1.3 = 11.418
Solve the equation.
log 3 𝑥 = 4
We can solve for x because we know that log 3 𝑥 = 4 means that 34 = 𝑥 = 81
4 − ln 5 − 𝑥 =
− ln 5 − 𝑥 = −
3
ln(5 − 𝑥) = 2
𝑒 ln(5 − 𝑥) = 𝑒
5−𝑥 =𝑒
3
2
3
2
−𝑥 = 𝑒 − 5
−𝑥 = −.5183
𝑥 = .5183
3
2
3
2
5
2
Use properties of logarithms to write the
expression as a logarithm of a single expression.
3
log 3 − log 3
1
3
1
2
First we can rewrite the expression as log 3 − log 3 and then with the properties it can be
1
written as a single expression as follows: log
1
2
1
3
1
6
32
1
33
and we can subtract exponents because
1
6
they are divided by each other − = so log 3 and then the exponent jumps out front
1
giving us: 6 log 3
Expand
4𝑥 3
log
.
𝑘
With the properties of logarithms, we can rewrite the expression as follows:
4𝑥 3
log 𝑘
= log 4𝑥 3 − log 𝑘 = log 4 + log 𝑥 3 − log 𝑘 = log 4 + 3 log 𝑥 − log 𝑘.
Solve the logarithmic equation.
ln 2𝑥 = 2
𝑒 ln 2𝑥 = 𝑒 2
2𝑥 = 𝑒 2
𝑥=
𝑒2
2
= 3.695
5 log 2 𝑥 = 25
5 log 2 𝑥 = 25
log 2 𝑥 5 = 25
We then can ask ourselves, 225 = 𝑥 5
33554432 = 𝑥 5
32 = 𝑥
Solve the logarithmic equation.
ln(5 − 𝑥) − ln 5 + 𝑥 = − ln 9
We can rewrite using properties of logarithms:
5−𝑥
ln 5+𝑥 = ln 9−1
5−𝑥
1
𝑒 ln 5+𝑥 = 𝑒 ln 9
5−𝑥
5+𝑥
1
=9
We can cross multiply and get: 9 5 − 𝑥 = 5 + 𝑥
45 − 9𝑥 = 5 + 𝑥
40 = 10𝑥
4=𝑥
The rest of the chapter review covers the
application problems. There will be application
problems on the exam so study accordingly.