Math 2250 Final Exam Practice Problem Solutions
1. What are the domain and range of the function
ln x
f (x) = √ ?
x
√
Answer: x is only defined for x ≥ 0, and ln x is only defined for x > 0. Hence, the domain of the
function is x > 0. Notice that
ln x
lim+ √ = −∞,
x
x→0
√
+
+
since x → 0 as x → 0 . Now, we can evaluate
ln x
lim √
x
x→∞
using L’Hôpital’s Rule; it is equal to
lim
x→∞
1
x
1
√
2 x
√
2 x
2
= lim
= lim √ = 0.
x→∞ x
x→∞
x
Therefore, f will have some maximum value; to figure out what it is, take
√ 1
1
2
√
x x − ln x 2√
− 2ln√xx
2 − ln x
x
2 x
0
f (x) =
=
=
.
x
x
2x3/2
Then f 0 (x) = 0 when
0 = 2 − ln x,
meaning that ln x = 2, or x = e2 . Notice that f 0 (x) changes sign from positive to negative at x = e2 ,
so the maximum of f occurs here. Since
2
ln e2
f (e2 ) = √ = ,
2
e
e
we see that the range of f is
2
.
−∞,
e
2. Find the inverse of the function f (x) = 1000(1 + 0.07)x .
Answer: To find the inverse, switch the roles of x and y, then solve for y:
x = 1000(1.07)y ;
taking the natural log of both sides, we see that
ln x = ln(1000(1.07)y ) = ln 1000 + ln(1.07y ) = ln 1000 + y ln 1.07.
Therefore,
y ln 1.07 = ln x − ln 1000.
Hence,
y=
ln x − ln 1000
.
ln 1.07
1
3. Find the point on the graph of y = e3x at which the tangent line passes through the origin.
Answer: Let f (x) = e3x . Since
f 0 (x) = 3e3x ,
the tangent line to e3x at the point x = a has slope 3e3a ; hence, using the point-slope formula, it is
given by
y − e3a = 3e3a (x − a) = 3e3a x − 3ae3a .
In other words, the tangent line to the curve at x = a is
y = 3e3a x − 3ae3a + e3a
or
y = e3a (3x − 3a + 1).
This passes through the origin if we get equality when we substitute 0 for both x and y, so it must be
the case that
0 = e3a (0 − 3a + 1) = e3a (1 − 3a).
Since e3a 6= 0, this means that 1 − 3a = 0, or a = 1/3. Therefore, since
f (1/3) = e3·1/3 = e,
the point whose tangent line passes through the origin is
1
,e .
3
4. Find the equation of the tangent line to the curve
xy 3 − x2 y = 6
at the point (3, 2).
Answer: Differentiating both sides with respect to x yields
y 3 + 3xy 2
dy
dy
− 2xy − x2
= 0.
dx
dx
Therefore,
dy
3xy 2 − x2 = 2xy − y 3 .
dx
Thus,
dy
2xy − y 3
=
.
dx
3xy 2 − x2
Plugging in (3, 2), we see that the slope of the tangent line is
2(3)(2) − 23
4
12 − 8
=
.
=
3(3)(2)2 − 32
36 − 9
27
Thus, using the point-slope formula, the equation of the tangent line is
y−2=
4
4
12
(x − 3) =
x− ,
27
27
27
or, equivalently,
y=
4
14
x+ .
27
9
2
√
5. Use an appropriate linearization to approximate 96.
√
√
Answer: Let f (x) = x. Then I will approximate 96 using the linearization of f at a = 100. To do
so, first take
1
f 0 (x) = √ .
2 x
Then the linearization is
1
x
x
(x − 100) = 10 +
−5=
+ 5.
20
20
20
L(x) = f (100) + f 0 (100)(x − 100) = 10 +
Therefore,
So we approximate
√
√
96 = f (96) ≈ L(96) = 5 +
48
98
96
=5+
=
= 9.8.
20
10
10
96 by 9.8.
2
6. Consider the function f (x) = x2 e−x . What is the absolute maximum of f (x)?
Answer: Notice that f is defined for all x. Also,
x2
2 = 0,
x→±∞ ex
lim f (x) = lim
x→±∞
(by two applications of L’Hôpital’s Rule) so f doesn’t go off to infinity.
Now, to find the critical points, compute
2
2
2
f 0 (x) = 2xe−x + x2 e−x (−2x) = e−x (2x − 2x3 ),
which equals zero precisely when 0 = 2x − 2x3 = 2x(1 − x2 ); namely when x = 0 or x = ±1 Thus, we
just need to evaluate f at the critical points:
f (1) = 1/e
f (0) = 0
f (−1) = 1/e
Since f limits to 0 in both directions, we see that the absolute maximum value of the function (occurring
at both x = 1 and x = −1) is 1/e.
7. A movie theater has been charging $7.50 per person and selling about 400 tickets on a typical weeknight.
After surveying their customers, the theater estimates that for every $1.50 that they lower the price,
the number of moviegoers will increase by 30 per night. This means the graph of the demand function
p(x) is a line passing through the points (400, 7.5) and (430, 6); using the point-slope formula, this
x
+ 27.5. Find the price which will maximize the theater’s revenue.
means that p(x) = − 20
Answer: Since revenue is given by the number of tickets sold (i.e. p(x)) times the price charged (i.e.
x), we know that the revenue function R(x) is
x
x2
R(x) = xp(x) = x − + 27.5 = − + 27.5x.
20
20
To maximize this, we need to find the critical points.
R0 (x) = −
so R0 (x) = 0 when
x
10
2x
x
+ 27.5 = − + 27.5,
20
10
= 27.5 or, equivalently, when x = 275.
3
The constraints on x are that 0 ≤ x ≤ 550 (since the theater would have to cut ticket prices to $0 to
get 500 customers), and the revenue for both of the endpoints is zero. Hence, the revenue is maximized
when x = 275. Now,
p(275) = −
27.5
27.5
275
+ 27.5 = −
+ 27.5 =
= 13.75
20
2
2
so the theater will maximize revenue when it charges $13.75 per ticket.
8. Water is draining from a conical tank at the rate of 18 cubic feet per minute. The tank has a height
of 10 feet and the radius at the top is 5 feet. How fast (in feet per minute) is the water level changing
2
when the depth is 6 feet? (Note: the volume of a cone of radius r and height h is πr3 h .)
Answer: If h is the height of the top of the water in the cone and r is the radius of the top of the
water, then
h
r
=
,
5
10
so r = h/2. Now, the volume of water in the tank is
V =
1 2
1
π 3
πr h = π(h/2)2 h =
h .
3
3
12
In turn, this means that
dV
π 2 dh
π dh
=
3h
= h2 .
dt
12
dt
4 dt
Since
dV
dt
= 18, this means that
18 =
π 2 dh
h
,
4 dt
or
dh
72
=
.
dt
πh2
Thus, when h = 6, the water level is changing at the rate
72
2
dh
=
= .
dt
36π
π
9. The function f (x) = x4 − 6x3 is concave down for what values of x?
Answer: To determine concavity, we need to compute the second derivative. Now,
f 0 (x) = 4x3 − 18x2 ,
so
f 00 (x) = 12x2 − 36x = 12x(x − 3).
Notice that f 00 (x) < 0 precisely when 0 < x < 3, so the function f is concave down on the interval
(0, 3).
10. Evaluate the limit
lim (1 − 6x)1/x .
x→0
Answer: Let f (x) − (1 − 6x)1/x . Taking the natural log of f yields
ln((1 − 6x)1/x ) =
1
ln(1 − 6x)
ln(1 − 6x) =
.
x
x
4
Now, by L’Hôpital’s Rule,
ln(1 − 6x)
= lim
x→0
x→0
x
lim ln(f (x)) = lim
x→0
−6
1−6x
1
= lim
x→0
−6
= −6.
1 − 6x
Therefore, since this is the limit of ln(f (x)), we know that
lim f (x) = e−6 .
x→0
11. Let f (x) = xcos x . What is f 0 (π/2)?
Answer: I will use logarithmic differentiation to find f 0 (x). To that end, let y = f (x) = xcos x . Then
ln y = ln(xcos x ) = cos x ln x.
Differentiating both sides,
1
cos x
1 dy
= cos x − sin x ln x =
− sin x ln x.
y dx
x
x
Therefore,
f 0 (x) =
cos x
cos x
dy
=y
− sin x ln x = xcos x
− sin x ln x .
dx
x
x
Hence,
f 0 (π/2) = (π/2)cos π/2
cos π/2
− sin π/2 ln(π/2)
π/2
= (π/2)0 (0 − 1 · ln(π/2)) = − ln(π/2).
12. For 0 ≤ t ≤ 5, a particle moves in a horizontal line with acceleration a(t) = 2t − 4 and initial velocity
v(0) = 3.
(a) When is the particle moving to the left?
Answer: The particle will be moving to the left when its velocity is negative. To determine the
velocity, note that
Z
Z
a(t)dt =
(2t − 4)dt = t2 − 4t + C.
Hence, v(t) = t2 − 4t + C for some C, which we can determine by plugging in t = 0:
3 = v(0) = 02 − 4(0) + C = C,
so v(t) = t2 − 4t + 3 = (t − 3)(t − 1). Notice that this function is negative when 1 < t < 3, so the
particle is moving to the left between t = 1 and t = 3.
(b) When is the particle speeding up?
Answer: The particle is speeding up when its acceleration is positive, which is to say when
0 < a(t) = 2t − 4,
so the particle is speeding up when t > 2.
(c) What is the position of the particle at time t if the initial position of the particle is 6?
Answer: Since
Z
Z
t3
v(t)dt = (t2 − 4t + 3)dt =
− 2t2 + 3t + D,
3
5
we know that s(t) =
in t = 0:
t3
3
− 2t2 + 3t + D for some real number D, which we can solve for by plugging
03
− 2(0)2 + 3(0) + D = D,
3
so the position of the particle at time t is
6 = s(0) =
s(t) =
13. If
R6
0
f (x)dx = 10 and
R4
0
f (x)dx = 7, find
R6
4
t3
− 2t2 + 3t + 6.
3
f (x)dx.
Answer: Notice that
6
Z
Z
4
6
Z
4
f (x)dx −
f (x)dx =
0
f (x)dx = 10 − 7 = 3.
0
14. Evaluate the definite integral
π/4
Z
sin tdt.
π/6
Answer: Since − cos t is an antiderivative of sin t, the Fundamental Theorem of Calculus tells us that
√
√
√
√
Z π/4
h
iπ/4
2
3
3− 2
sin tdt = − cos t
= − cos(π/4) − (− cos(π/6)) = −
+
=
.
2
2
2
π/6
π/6
15. Evaluate the integral
Z
2
dt.
t−3
2
Answer: Since t−3
looks vaguely like 1t , we should expect that the natural log comes into play. In
2
, so
fact, 2 ln(t − 3) is an antiderivative of t−3
Z
2
dt = 2 ln(t − 3) + C.
t−3
16. Evaluate the definite integral
Z
1
4
√
2 x + 4x2
dx
x
Answer: Re-write the integral as
Z 4 √
Z 4 √
Z 4 2
Z 4
Z 4
2 x 4x2
2 x
4x
2
√ dx +
+
dx =
dx +
dx =
4xdx.
x
x
x
x
x
1
1
1
1
1
Now,
Z
4
2
√ dx =
x
1
Z
4
2x−1/2 dx =
1
2x1/2
1/2
4
1
√ 4
= 4 x 1 = 8 − 4 = 4.
On the other hand,
Z
1
Therefore,
Z
1
4
4
4
4xdx = 2x2 1 = 32 − 2 = 30.
√
Z 4
Z 4
2 x + 4x2
2
√ dx +
dx =
4xdx = 4 + 30 = 34.
x
x
1
1
6
17. Suppose the velocity of a particle is given by
v(t) = 6t2 − 4t.
What is the displacement of the particle from 0 to 2?
Answer: The displacement is given by
s(2) − s(0).
0
Since s (t) = v(t), the Fundamental Theorem tells us that
Z
2
s(2) − s(0) =
Z
0
2
v(t)dt =
s (t)dt =
0
0
0
2
Z
2
(6t2 − 4t)dt = 2t3 − 2t2 0 = (16 − 8) − (0 − 0) = 8.
Therefore, the displacement is 8 units.
18. Suppose that
Z
x2
f (t)dt =
p
x2 + 1 − 1.
0
What is f (2)?
Answer: Let g(x) =
√
x2 + 1 − 1. Then,
d
g (x) =
dx
0
x2
Z
!
f (t)dt
0
d
=
du
Z
0
u
du
f (t)dt
dx
where u = x2 , using the Chain Rule.
Therefore, by the first part of the Fundamental Theorem,
g 0 (x) = f (u) · 2x = 2xf (x2 ).
In other words,
f (x2 ) =
Now, we know that g(x) =
√
g 0 (x)
.
2x
x2 + 1 − 1, so
1
x
g 0 (x) = √
(2x) = √
.
2
2
2 x +1
x +1
Hence,
√
√
√2
g 0 ( 2)
1
f (2) = √ = √3 = √ .
2 2
2 2
2 3
19. Evaluate the integral
Z
3etan x sec2 x dx.
Answer: Let u = tan x. Then du = sec2 x dx, so we can re-write the above integral as
Z
3eu du = 3eu + C.
Now, substituting back in for u, this yields the answer
3etan x + C.
7
20. Evaluate the definite integral
π/16
Z
8 tan(4x) dx.
0
Answer: Notice that tan(4x) =
sin(4x)
cos(4x) ,
so the above integral is equal to
Z
π/16
sin(4x)
dx.
cos(4x)
8
0
Let u = cos(4x). Then du = −4 sin(4x) dx. Therefore,
8
sin(4x)
−2
−2
dx =
· (−4 sin(4x)dx) =
du.
cos(4x)
cos(4x)
u
We want to replace the given integral with an integral in terms of u, but that means we also need to
change the limits of integration:
u(0) = cos(4 · 0) = cos(0) = 1
1
u(π/16) = cos(4 · π/16) = cos(π/4) = √ .
2
and
Therefore, the integral we were given can be re-written as
Z
1
√
1/ 2
√
√
√
−2
1/ 2
du = [−2 ln u]1
= −2 ln(1/ 2) + 2 ln(1) = −2 ln(1/ 2) = −2 ln 2−1/2 .
u
But now, from the properties of logarithms,
−1/2
−2 ln 2
Hence, we conclude that
R π/16
0
= ln
−1/2
2
−2 = ln 21 = ln 2.
8 tan(4x) dx = ln 2.
21. What is the area of the red region in the figure? The blue curve is given by y = 2 lnxx and the vertical
green lines are the lines x = 1e and x = e.
Answer: First, notice that the curve y = 2 lnxx crosses the x-axis when
2
ln x
= 0,
x
8
which only occurs when ln x = 0, meaning when x = 1. Between x = 1/e and x = 1, the x-axis (i.e.
the curve y = 0) is above the blue curve, whereas the blue curve is above the x-axis when 1 < x ≤ e.
Therefore, the red area is equal to
Z e
Z 1
Z e
Z 1 ln x
ln x
ln x
ln x
2
−2
2
0−2
dx +
− 0 dx =
dx +
dx.
x
x
x
x
1
1/e
1
1/e
For both of these integrals, let u = ln x. Then du =
1
x
dx. Moreover,
u(1/e) = ln(1/e) = ln(e−1 ) = −1
u(1) = ln(1) = ln(e0 ) = 0
u(e) = ln(e) = ln(e1 ) = 1.
Therefore, the sum of integrals above is equal to
Z 1
Z 0
1
0
2u du = −u2 −1 + u2 0
−2u du +
0
−1
= −(0)2 + (−1)2 + 12 − 02
= 2,
so the red region has area 2.
9