Engineering Math I – Fall 2014
Quiz #8 Solutions
Problem 1 (4 pts). If g is the inverse function of f (x) = 2x + ln x, find g 0 (2).
If g is the inverse of f , the g(f (x)) = x for all x. We will use the identity g 0 (x) =
1/f 0 (g(x)), so we need to find g(2) and f 0 (g(2)). First, g(2) is the unique x such that
f (x) = 2, so we solve 2x + ln x = 2, which upon inspection gives x = 1. So g(2) = 1. Now
we find f 0 (1). Differentiating f (x) = 2x + ln x with respect to x yields f 0 (x) = 2 + x1 , and
thus f 0 (1) = 3. Thus g 0 (2) = 13 , so the correct answer is (d).
Problem 2 (4 pts). Find limx→∞ tan−1 (x2 ).
Solution. The limit is
correct answer is (b).
π
,
2
as is easily seen in the graph of y = tan−1 (x) below. So the
Problem 3 (6 pts). Let y = ln ln x.
(a) (2 pts) Find the domain of y.
Solution. We must have ln x > 0, which means that eln x > e0 , which implies that
x > 1. So the domain is (1, ∞).
(b) (2 pts) Find y 0 .
Solution. The derivative of ln x is x1 , so by the chain rule, the derivative of y =
ln ln x is y 0 = x ln1 x .
(c) (2 pt) Find an equation of the tangent line to y at the point (e, 0).
Solution. The slope of the tangent line is given by y 0 (e) =
point-slope form, we have y = 1e (x − e) = xe − 1.
1
e ln e
= 1e . Using the
MATH 151:549-551 – Fall 2014 Quiz #8 Solutions
2
Problem 4 (6 pts). A certain object cools at a rate (in ◦ C/min) equal to one-tenth of
the difference between its temperature and that of the surrounding air. If a room is kept
at 21◦ C and the temperature of the object is 33◦ C, find an expression for the temperature
of the object t minutes later.
Solution. Let y(t) be the temperature of the object. Since the temperature of the room
(i.e. temperature of the surrounding air) is kept at 21◦ C, we have
dy
1
=
y(t) − 21 .
dt
10
1
1
Let u(t) = y(t) − 21. Then we may write dy
= 10
u. Taking k = 10
, we may write the
dt
kt
expression for the temperature u(t) = u(0)e , where u(0) = y(0) − 21 = 33 − 21 = 12.
t
Thus u(t) = 12e 10 . Since u(t) = y(t) − 21, we see that
t
y(t) = 21 + 12e 10 .
Bonus Problem (2 pts). A curve passes through the point (0, 5) and has the property
that the slope of the tangent line to the curve at every point P is twice the y-coordinate
of P . What is the equation of the curve?
Solution. Let y(t) be the equation of the curve. The slope of the tangent line to the curve
at any point is given by y 0 (t), and the y-coordinate of any point P on the curve is just
y(t). Thus, y 0 (t) = 2y(t). So the equation of the curve can be expressed as y(t) = y(0)e2t .
We are told that the curve passes through the point (0, 5), and thus y(0) = 5. Therefore,
y(t) = 5e2t is the equation of the curve.