MATH A105 Intermediate Algebra
Exam 3 Key
Instructions
1. Do NOT write your answers on these sheets. Nothing written on the test papers will be graded.
2. Do NOT write your name on any of your answer sheets.
3. Please begin each section of questions on a new sheet of paper.
4. Do not write problems side by side.
5. Do not staple test papers.
6. Limited credit will be given for incomplete or incorrect justification.
Questions
1. Lines and Exponentials
(a) (3) Find the slope of the line 7x − 3y = 9.
x
=
0.
0 − 3y
=
9.
y
x
14 − 3y
−3y
y
m
= −3.
=
=
2.
9.
= −5.
=
5/3.
5/3 − (−3)
=
2−0
14/3
=
2
= 7/3.
or
7x − 3y
=
−3y
9.
= −7x + 9.
1
= − (−7x + 9).
3
7
=
x − 3.
3
= 7/3.
y
y
m
(b) (5) Write an equation of the line through (0, −3) and (3, 4).
m
=
=
y−4
1
=
4 − (−3)
3−0
7
.
3
7
(x − 3).
3
(c) (2) Rewrite log3 x = 4 as an exponential.
34 = x.
(d) (3) Solve 4 + log2 (x + 5) = 7.
4 + log2 (x + 5)
=
7.
−4 + 4 + log2 (x + 5)
=
−4 + 7.
log2 (x + 5)
=
3.
3
= x + 5.
8
= x + 5.
2
8−5
3
= x + 5−5.
= x.
(e) (3) Solve 7(ex−4 ) = 7.
7(ex−4 ) = 7.
7(ex−4 )
7
=
.
7
7
ex−4 = 1.
ln(ex−4 ) = ln(1).
x−4
=
0.
x−4+4
=
0 + 4.
x =
2
4.
2. Functions
f (x) = 5x2 − 6. g(x) = 4x − 1. h(x) = 4 + log2 (x + 5).
(a) (3) Evaluate each of the following. Expand, collect, reduce as appropriate.
i. f (2)
f (2)
=
5(2)2 − 6
=
5(4) − 6
=
20 − 6
=
14.
=
4(a − 4) − 1
=
4a − 16 − 1
=
4a − 17.
ii. g(a − 4)
g(a − 4)
iii. f (g(x))
f (g(x))
= f (4x − 1)
= 5(4x − 1)2 − 6
=
5(16x2 − 8x + 1) − 6
=
80x2 − 40x + 5 − 6
=
80x2 − 40x − 1.
(b) (3) Find g −1 (x).
g(x)
=
4x − 1.
g(x)+1
=
4x − 1+1.
g(x) + 1
g(x) + 1
4
g(x) + 1
4
=
g −1 (x)
=
4x.
4x
=
.
4
= x.
x+1
.
4
(c) (3) Find h−1 (x).
h(x)
=
4 + log2 (x + 5).
−4 + h(x)
= −4 + 4 + log2 (x + 5).
−4 + h(x)
=
−4+h(x)
2
2−4+h(x) −5
2−4+h(x) − 5
−1
h
(x)
log2 (x + 5).
= x + 5.
= x + 5−5.
= x.
=
3
2−4+x − 5.
(d) (3) Does f (x) have an inverse function?
f (x)
=
5x2 − 6.
f (x) + 6
f (x) + 6
5
r
f (x) + 6
5
=
5x2 .
= x2 .
= |x|.
Each output has two inputs, so the inverse is not a function.
4
3. Applications and Data
(a) The half-life of Cesium is about 30.17 years. This means that every 30.17 years half of the cesium decays
t/30.17
into something else. This can be expressed as C(t) = C0 21
. (2 each)
i. If you start with 4 grams of Cesium. How much will you have after 60.34 years?
60.34/30.17
1
C(60.34) = 4
2
2
1
= 4
2
1
= 4·
4
= 1.
ii. Why is the base (number in parentheses) 1/2?
This is the half-life: every time is cuts in half.
iii. What role does the division by 30.17 play?
This is the length of the half-life. Dividing by it determines how many half-lives have passed.
iv. What is C0 ?
This is the initial amount.
(b) Determine whether each set varies linearly, exponentially, or otherwise. (2 each)
i.
x
y
−
÷
1
2
2
3
4
5
3 9/2 27/4 81/8
1 3/2 9/4 27/8
3/2 3/2 3/2 3/2
x
y
−
1
2
2
3
4
5
9/2 14/2 19/2 12
5/2 5/2 5/2 5/2
This is exponential.
ii.
This is linear.
iii.
x
1
y −6
−
÷
2
−1
5
1/6
3
4
5
14
39
74
15
25
35
−14 39/14 74/39
This is something else.
iv.
x 1
y 6
−
2
3
4
1 −4 −9
−5 −5 −5
This is linear.
5
5
−14
−5