Engineering Math I – Fall 2014
Quiz #7 Solutions
Problem 1 (4 pts). Let f (x) = sin x. Which of the following is true?
(a) f is one-to-one.
(b) The function g = sin−1 x is the inverse of f for all x.
(c) f is one-to-one for − π2 ≤ x ≤
π
2
(d) f is one-to-one for 0 ≤ x ≤ π.
Solution. The correct answer is (c). The sin function does not pass the horizontal line
test, so it is not one-to-one. Thus the function sin−1 cannot be the inverse of sin for all
x. When the domain is restricted to 0 ≤ x ≤ π, the function still does not pass the
horizontal line test. These facts eliminate (a), (b) and (d).
Problem 2 (4 pts). Suppose g is the inverse function of f and f (4) = 5, f 0 (4) = 32 . Find
g 0 (5).
Solution. If g is the inverse of f , then g(f (x)) = x for all x. Thus g(5) = g(f (4)) = 4.
Using the identity g 0 (x) = 1/f 0 (g(x)), we have
g 0 (5) =
1
f 0 (g(5))
=
1
f 0 (4)
=
1
2
3
3
= .
2
So the correct answer is (c).
Problem 3 (5 pts). The geologist C. F. Richter defined the magnitude of an earthquake
to be log10 (I/S), where I is the intensity of the quake (measured by the amplitude
of a seismograph 100 km from the earthquake) and S is the intensity of a “standard”
earthquake (where the amplitude is only 1 micron = 10−4 cm). The 1989 Loma Prieta
earthquake that shook San Francisco had a magnitude of 7.1 on the Richter scale. The
1906 San Francisco earthquake was 16 times as intense. What was its magnitude on the
Richter scale?
Solution. Let I be the intensity of the Loma Prieta earthquake in 1989. We are told
in the problem that log10 (I/S) = 7.1. We are also told that intensity of the 1906 San
Francisco earthquake was 16I. So the magnitude of the 1906 earthquake was
log10 (16I/S) = log10 (16) + log10 (I/S) = log10 (16) + 7.1.
MATH 151:549-551 – Fall 2014 Quiz #7 Solutions
2
The following problem had a typo when given in class, which has been fixed on this
version.
2
Problem 4 (7 pts). Let r(t) = h √tt+3 , ln(t2 − 1)i
(a) (2 pts) What is the domain of r(t)?
√
t
Solution. The domain of √t+3
is (−3, ∞), since t + 3 must be strictly greater
than 0. The domain of ln(t2 −1) is (−∞, −1)∪(1, ∞) since we must have t2 −1 > 0
which implies that t2 > 1 and after taking the square root gives |t| > 1. Thus the
domain of r(t) is (−3, −1) ∪ (1, ∞).
(b) (3 pts) Does the curve traced by r(t) in the Cartesian plane ever cross the x-axis?
What about the y-axis? Determine the quadrant(s) in which the curve lies.
Solution. The curve crosses the x-axis when y = 0, which
means that ln(t2 −1) = 0.
p
This occurs when t2 − 1 = 1, which means that t = ± (2). Both of these values
for t are in the domain, so the curve does cross the x-axis.
2
The curve crosses the y-axis when x = 0, which means that √tt+3 = 0. This occurs
when t = 0, but this is not in the domain of the function, so the curve does not
cross the y-axis.
Since the curve does not cross the y-axis, it must lie in either the first and third
quadrant, or the second and fourth quadrant. The x-coordinate of any point on
2
the curve is given by √tt+3 , which is always positive, so the curve lies in the first
and third quadrants.
(c) (2 pts) For which value(s) of t does r(t) = h4, ln 3i?
Solution. First, note that ln(t2 − 1) = ln 3 if and only if t2 − 1 = 3. That is, t2 = 4,
2
which happens when t = ±2. Plugging in 2 and −2 into √tt+3 , we see that t = −2
is the correct value.