Math 122 – Quiz IV Revue
1. Express the statement as an equation, then find the constant of proportionality, k.
a) z varies inversely as t. If t = 3, then z = 5.
z =
k
t
5 =
k
3
15 = k
b) S varies jointly as p and q. If p = 4, and q = 5, then S = 180.
S = kpq
180 = k(4)(5)
180 = 20k
9 = k
c) W is inversely proportional to the square of r. If r = 6 then W = 10.
W
=
k
r2
10 =
k
(6)2
10 =
k
36
360 = k
2
d) t is jointly proportional to x and y, and inversely proportional to r. If x = −1, y = 4
, and r = −2, then t = 14
t = k
xy
r
14 = k
(−1)(4)
−2
14 = k
−4
−2
14 = k(2)
7 = k
2. The loudness L of a sound (measured in decibels, dB) is inversely proportional to the
square of the distance d from the source of the sound. A person who is 10 ft from a lawn
mower experiences a sound level of 70 dB. How loud is the lawn mower when the person
is 100 ft away?
k
First we need our equation, which is L = 2 . Then we plug in d = 10 and L = 70, to
d
solve for k as follows:
k
L =
d2
70 =
k
(10)2
70 =
k
100
7000 = k
Now that we have k, we can rewrite our equation with the value of k, and plug in
d = 100.
7000
L =
d2
L =
7000
(100)2
L = 0.7 dV
3
3. For each function, (i) find the domain, (ii) graph the function, and (iii) evaluate the
function at x = −2, 0, 4.
a) f (x) =
1
x−5
(i) Here, we can’t have the denominator equal zero, so we need to look at where x−5 6= 0,
so x 6= 5. Thus,
D : (−∞, 5) ∪ (5, ∞)
(ii)
1
1
=−
−2 − 5
7
1
1
f (0) =
=−
0−5
5
1
1
f (2) =
=−
2−5
3
(iii) f (−2) =
b) g(x) = x2 + x + 8
(i) Since this is a polynomial, with no square root, and no fraction, we don’t have any
problems for x. Thus
D : (−∞, ∞)
(iii) g(−2) = (−2)2 + (−2) + 8 = 4 − 2 + 8 = 10
g(0) = (0)2 + (0) + 8 = 8
g(2) = (2)2 + (2) + 8 = 4 + 2 + 8 = 14
4
(ii)
c) h(x) =
√
x−3
(i) Since we have a square root, we need the stuff inside the square root to be greater
than or equal to zero. So we need x − 3 ≥ 0, or x ≥ 3. Therefore, we have
D : [3, ∞)
(ii)
(iii) h(−2) =
h(0) =
h(2) =
√
√
p
(−2) − 3 =
0−3=
2−3=
√
√
√
−5
−3
DNE
−1
DNE
DNE
I didn’t mean for all of these to not exist...sorry it worked out like that...that won’t
happen all the time.
5
4. Let f (x) = x2 + 1. Find each of the following and simplify if possible.
(a)
f (a) = (a)2 + 1
= a2 + 1
(b)
f (a + h) = (a + h)2 + 1
= a2 + 2ah + h2 + 1
(c)
f (a + h) − f (a)
h
=
(a2 + 2ah + h2 + 1) − (a2 + 1)
h
=
2ah + h2
h
=
h(2a + h)
h
= 2a + h