COLLEGE ALGEBRA I (MATH 006)
SPRING 2016
——PRACTICE FOR EXAM IV——
——SOLUTIONS——
1. Suppose that the relation S is defined as follows.
S = {(9, −8), (0, 9), (−7, 4), (−7, 5)}
Give the domain and range of S using set notation.
Is S a function?
Solution:
Domain= {9, 0, −7}, Range= {−8, 9, 4, 5, }
Not a function:{(-7,4), (-7,5) }
2. Express the statement as an equation, and find the constant
of proportionality.
(a) y is directly proportional to x. If x = 6, then y = 42.
(b) R is inversely proportional to s. If s = 4, then R = 3.
(c) F varies jointly as the square of c and d. If c = 30 and
d = 20, then F = 150.
(d) M varies directly as x and inversely as y. If x = 2 and
y = 6, then M = 5.
Solution:
a) y = kx
b) R = ks
42 = 6k
3 = k4
k=7
k = 12
y = 7x
R = 12
s 2
c) F = kc d
d) M = k xy
150 = k(900)(20)
5 = k 26
1
k = 120
k = 15
c2 d
F = 120
M = 15 xy
1
3. The function T is one-to-one. Find T −1 and state its domain and range using set notation.
T = {(−9, −2), (−2, 7), (2, −4), (4, 6)}
Solution:
T −1 = {(−2, −9), (7, −2), (−4, 2), (6, 4)}
Domain= {−2, 7, −4, 6}, Range= {−9, −2, 2, 4}
4. Show that f and g are inverses of each other by calculating
f (g(x)) and g(f (x)).
f (x) = 3x − 1, g(x) = x+1
3
Solution:
f (g(x)) = 3g(x) − 1 = 3( x+1
3 )−1 =x+1−1 =x
g(f (x)) = f (x)+1
= 3x−1+1
= 3x
3
3
x =x
5. The following function is one-to-one. Find f −1 .
7x
f (x) = 2x−1
Solution:
7x
y = 2x−1
7y
x = 2y−1
x(2y − 1) = 7y
2xy − x = 7y
2xy − 7y = x
y(2x − 7) = x
x
y = 2x−7
x
f −1 (x) = 2x−7
2
6. The following
function is one-to-one. Find h−1 .
√
h(x) = x − 2, x ≥ 2
Solution:
√
y = x−2
√
x= y−2
√
x2 = ( y − 2)2
x2 = y − 2
y = x2 + 2
h−1 (x) = x2 + 2
7. A hot bowl of soup starts to cool according to Newton’s
Law of Cooling. Its temperature at time t is given by
T (t) = 65 + 145e−0.05t
where t is measured in minutes and T is measured in o F .
(a) What is the initial temperature of the soup?
(b) What is the temperature after 10 minutes? (Answer
does not have to be exact.)
Solution:
(a) T (0) = 65 + 145e−0.05(0) = 210
(b) T (10) = 65 + 145e−0.05(10) = 152.9
8. Rewrite as an exponential equation.
1
(a) log3 81
= −4
(b) ln 5 = y
Solution:
1
(a) 3−4 = 81
(b) ey = 5
3
9. Solve for x.
log7 x = −2
Solution:
1
x = 7−2 = 49
√
2
x +1
10. Use the properties of logs to expand log x3 (x+1)
4 , x > 0.
Solution:
√
h
i
√
x2 +1
2 + 1 − log x3 (x + 1)4
=
log
x
log x3 (x+1)
4
= 21 log(x2 + 1) − [3 log x + 4 log(x + 1)]
11. Write as a single log.
1
3 log w + 5 log x − log z
Solution:
1
log w 3 + log x5 − log z
1
= log(w 3 x5 ) − log z
1
= log w 3zx
5
12. Solve for x.
log5 (x − 3) = 1 − log5 (x − 7)
Solution:
log5 (x − 3) + log5 (x − 7) = 1
log5 [(x − 3)(x − 7)] = 1
log5 (x2 − 10x + 21) = 1
x2 − 10x + 21 = 51
x2 − 10x + 16 = 0
(x − 8)(x − 2) = 0
x = 8 or x = 2(Extraneous- negative log argument)
x=8
4
13. Solve for x. e−8x = 6
Solution:
ln e−8x = ln 6
−8x ln e = ln 6
−8x = ln 6
x = − ln86
x = −.224
14. A sum of $100 is invested at an interest rate of 8% per year.
Find the time required for the money to grow to $150 if the
interest is compounded
(a) semi-annually
(b) continuously.
Solution:
2t
(a) 150 = 100 1 + .08
2
2t
3
=
(1.04)
2
ln 23 = ln (1.04)2t
ln 23 = 2t ln (1.04)
ln 32
t = 2 ln(1.04)
t ≈ 5.17yrs
(b) 150 =
3
2 =
ln 32 =
ln 23 =
t=
t≈
100e.08t
e.08t
ln e.08t
.08t
ln 32
.08
5.07yrs
3
15. Find the domain of the function. f (x) = log x−4
Solution:
3
x−4 > 0
x > 4 or (4, ∞)
5