Review Problems
(largely taken from old midterms)
Problem 1
Find the following limits or find that the limit does not exist (you may NOT use
l’Hopital’s rule)
a.
tan(5x)
sin(5x)
1
= lim 5
x→0
x→0
x
5x cos(5x)
1
=5·1·
1
=5
lim
b.
|x − 2|
x→2 x − 2
Calculate the one sided limits separately:
|x − 2|
x−2
lim
= lim
=1
x→2+ x − 2
x→2+ x − 2
and
|x − 2|
−(x − 2)
lim
= lim
= −1
x→2− x − 2
x→2−
x−2
lim
1
2
Problem 2
Find A and B such that the following function is continuous
 2
x≤0
 x
Ax
+
B
0
<
x
≤ 25
f (x) =
 √x−5
25 < x
x−25
Use the definition of continuity applied at x = 0 and x = 25. In other words find four
one-sided limits and equate the appropriate pairs to get equations for A and B:
lim f (x) = lim x2 = 0
x→0−
x→0−
lim f (x) = lim Ax + B = B
x→0+
x→0+
lim f (x) = lim Ax + B = 25A + B
x→25−
√
√
x − 25 1
1
x−5 x+5
√
√
= lim
=
lim f (x) = lim
x→25+
x→25+ x − 25
10
x + 5 x→25+ x − 25 x + 5
And so you get the two equations
x→25−
0=B
1
25A + B =
10
which give A =
1
250
and B = 0.
3
Problem 3
The function f (x) is continuous everywhere and it has the values f (−3) = −1, f (−1) =
0, f (0) = 1, f (3) = 2, f (5) = 0, f (6) = 1, f (8) = −2. What can you say about the
number and location of solutions to the equation f (x) = 0? State any theorems that you
are using.
The following picture indicates the answer, that there are at least three solutions, at
x = −1, x = 5 and one in the interval (6, 8) according to the Intermediate Value Theorem.
Problem 4
Find the derivative of f (x) = x1 USING ONLY THE DEFINITION OF THE DERIVATIVE.
1
− x1 (x + h)(x)
f (x + h) − f (x)
x+h
′
= lim
f (x) ≡ lim
h→0
h→0
h
h
(x + h)(x)
x − (x + h)
= lim
h→0 h(x + h)(x)
−h
= lim
h→0 h(x + h)(x)
−1
= lim
h→0 (x + h)(x)
−1
= 2
x
4
Problem 5
Find the derivatives of the following functions:
a.
f (x) = sin(ex )x4.2
Use the product and chain rules to get
f ′ (x) = cos(ex )ex x4.2 + 4.2 sin(ex )x3.2 .
b.
y = xsin(x)
Use logarithmic differentiation as follows.
ln(y) = ln xsin(x)
= x sin(x),
so that implicitly differentiating with respect to x we get
1 dy
= sin(x) + x cos(x).
y dx
Then solving for
dy
:
dx
dy
= y(sin(x) + x cos(x))
dx
and substituting for y:
dy
= xsin(x) (sin(x) + x cos(x))
dx
5
Problem 6
Find the tangent line to
through the point (2, −4).
Use implicit differentiation:
3x3 + 2y 2 − 2x + 8y = 20
9x2 + 4y
and solving for
dy
dy
−2+8
=0
dx
dx
dy
:
dx
dy
dy
2 − 9x2
2
(4y + 8) = 2 − 9x ⇒
=
dx
dx
4y + 8
so the slope of the tangent line is
2 − 36
−34
17
2 − 9 · 22
=
=
= .
4 · −4 + 8
−16 + 8
−8
4
then using the slope form of the line we get that the tangent is:
17
y+4
=
4
x−2
6
Problem 7
Consider
p(x) = 2x3 − 9x2 − 60x + 4.
a) Find the intervals of increase and decrease of p(x).
p′ (x) = 6x2 − 18x − 60 = 6(x2 − 3x − 10) = 6(x + 2)(x − 5)
so the critical values are x = −2, 5 and a sign analysis of p′ (x) yields
+ + + + + + +(−2) − − − − − − − (5) + + + + + + + +
So p(x) is increasing when x < −2 and x > 5 and decreasing for −2 < x < 5
b) Find the intervals of concavity of p(x)
p′′ (x) = 12x − 18
so that the only 2nd order critical value is x = 3/2 and a sign analysis yields
− − − − − − − − (3/2) + + + + + + + +
so that p(x) is concave up when x > 3/2 and concave down for x < 3/2
c) Find every relative minimum and maximum of p(x). And state which test you are
using and why it works.
Relative Maximum: x = −2 either 1st (p′ (x) switches from positive to negative around
x = −2) or 2nd (p′′ (−2) < 0) derivative tests
Relative Minimum: x = 5 either 1st (p′ (x) switches from negative to positive around
x = 5) or 2nd (p′′ (5) > 0) derivative tests
d) Find every inflection point of p(x).
Inflection point: x = 3/2 (the sign of the second derivative changes around x = 3/2)
e) Sketch the graph of p(x).
7
Problem 8
Sketch the graph of a function f (x) satisfying the following conditions:
• f (x) has horizontal asymptotes at y = −3, 4.
• f (x) has vertical asymptotes at x = −1, 2.
• f (x) has a cusp at x = 0.
• f (x) is defined and continuous everywhere except at x = −1, 2 and is differentiable
everywhere except at x = −1, 0, 2.
• f (x) has a relative minimum at x = 4.
• f (x) has exactly one inflection point for x > 2.
Note: There is more than one correct answer.
8
Problem 9
Find all horizontal and vertical asymptotes of
(x − 3)(x − 4)(x − 1)
h(x) =
(x + 1)(x − 2)(x + 4)
There are vertical asymptotes at x = −1, 2, −4.
(x − 3)(x2 − 5x + 4)
(x + 1)(x2 + 2x − 8)
x3 − 5x2 + 4x − 3x2 + 15x − 12
=
x3 + 2x2 − 8x + x2 + 2x − 8
x3 − 8x2 + 19x − 12
= 3
x + 3x2 − 6x − 8
h(x) =
So
x3 − 8x2 + 19x − 12 1/x3
lim h(x) = lim
x→±∞
x→±∞ x3 + 3x2 − 6x − 8
1/x3
1 − 8/x + 19/x2 − 12/x3
= lim
x→±∞ 1 + 3/x − 6/x2 − 8/x3
=1
So that h(x) has one horizontal asymptote y = 1.
(Alternatively you can use l’Hopital’s rule to evaluate the limit.)
9
Problem 10
Determine whether the following functions have a cusp, a vertical tangent, both or
neither:
4
f (x) = x 5 (x − 1)
3
g(x) = x 5 (x − 1)
so
f (x) = x9/5 − x4/5
4
9
f ′ (x) = x4/5 − x−1/5
5
5
−1/5
x
=
(9x − 4)
5
1
= 1/5 (9x − 4)
5x
and so f ′ (x) has an asymptote at x = 0 where f (x) is defined. Further
lim f ′ (x) = −∞
x→0+
lim f ′ (x) = +∞
x→0−
and since the signs are opposite f (x) has a cusp.
so
g(x) = x8/5 − x3/5
8
3
g ′ (x) = x3/5 − x−2/5
5
5
x−2/5
=
(8x − 3)
5
1
= 2/5 (8x − 3)
5x
and so g ′ (x) has an asymptote at x = 0 where g(x) is defined. Further
lim g ′ (x) = −∞
x→0+
lim g ′ (x) = −∞
x→0−
and since the signs are the same g(x) has a vertical tangent.
10
Problem 11
Use a linear approximation to estimate
√
3
8.1
Let
f (x) =
so
p
3
(x) = x1/3
1
f ′ (x) = x−2/3
3
and the linear approximation to f (x) at x = 8 is
L(x) = f (8) + f ′ (8)(x − 8)
p
1
= 3 (8) + (8)−2/3 (x − 8)
3
1
(x − 8)
=2+
3 · 82/3
1
= 2 + (x − 8)
12
and then
√
3
8.1 ≃ L(8.1) = 2 +
1
0.1
1
(8.1 − 8) = 2 +
=2+
12
12
120
11
Problem 12
Consider the following function
q(x) = x3 − 2x2 − 5x + 6.
You would like to estimate a root and you decide to use the Newton-Raphson method.
You begin at x = −1. Find the next two iterations that the method yields (you do NOT
need to simplify the second iterate).
Recall the iterative formula
f (xn )
xn+1 = xn − ′
f (xn )
and also
q ′ (x) = 3x2 − 4x − 5.
We begin with x0 = −1.
so
q(−1) = −1 − 2 + 5 + 6 = 8
q ′ (−1) = 3 + 4 − 5 = 2.
x1 = −1 −
8
= −5
2
And continuing one more step
q(−5) = −125 − 2(25) + 25 + 6 = −144
so
q ′ (−5) = 75 + 20 − 5 = 90
x2 = −5 −
−144
.
90
12
Problem 13
You are measuring the volume of a cylinder. You know that it is exactly 4 inches tall.
By wrapping a string around the outside of the cylinder, you find that the circumference
is 5 inches, give or take 0.2 in.
a) Find the percentage error in the circumference measurement.
Since
c=5
dc = 0.2
therefore the percentage error in the circumference is
0.2
20
dc
100% =
100% = % = 4%.
c
5
5
b) Recalling the formula
1 2
ch
V =
4π
for the volume of a cylinder in terms of its circumference c and height h, find the calculated volume of the cylinder.
25
1
(5)2 (4) = in3
4π
π
c) Find the percentage error in the calculated volume.
We have
2h
dV =
cdc
4π
so that
2h
cdc
dV
2dc
= 4π
1 2 =
V
c
ch
4π
V =
and thus
∆V
dV
dc
100% ≃
100 = 2 100 = 2(4) = 8%.
V
V
c
13
Problem 14
A particle is moving along a line according to the following position function
s(t) = t3 − 6t2 + 9t + 1.
Calculate both the total distance traveled and also the (signed) displacement between
t = 0 and t = 4.
displacement = s(4) − s(0) = 5 − 1 = 4.
ds(t)
= v(t) = 3t2 − 12t + 9
dt
= 3(t2 − 4t + 3)
= 3(t − 1)(t − 3).
So the turnaround times are t = 1, 3. Further
s(0) = 1
s(1) = 1 − 6 + 9 + 1 = 5
s(3) = 27 − 54 + 27 + 1 = 1
Therefore
s(4) = 64 − 6(16) + 36 + 1 = 101 − 96 = 5.
distance = |s(1) − s(0)| + |s(3) − s(1)| + |s(4) − s(3)|
= |5 − 1| + |1 − 5| + |5 − 1| = 12.
14
Problem 15
You are flying a kite. It is drifting in a straight line up and away from you. It is moving
2
more and more slowly away from you horizontally with speed e−t ft/s and it is rising at
3 ft/s. Assume that the kite starts on the ground some distance away from you and that
after 10 seconds, the kite is 40 ft distant from you horizontally. Find out how fast the
string is unwinding at that moment.
Given:
dx
2 dy
= e−t
= 3x(10) = 40
dt
dt
We have
l 2 = x2 + y 2
so that
2l
dl
dx
dy
= 2x + 2y
dt
dt
dt
so that
x dx + y dy
dl
dt
= dt
dt
l
(0.1)
and at t = 10 y = 30 so that
l(10)2 = 402 + 302
and so l(10) = 50 (3 − 4 − 5 triangle scaled). Then plugging in values for t = 10 into 0.1,
we get
dl
40e−100 + 30 · 3
|t=10 =
=
dt
50
4 −100 9
e
+
5
5
ft/s
15
Problem 16
You are building a fence in the partitioned rectangular shape as pictured below. The
total length of fencing you have is 600 ft. Find x and y which enclose the largest area.
We want to maximize the area, which is
A = 3xy
where
4x + 6y = 600
so that
3y = 300 − 2x
and the area in terms of x alone then becomes
A(x) = x(300 − 2x) = −2x2 + 300x
which function we would like to maximize on the interval 0 ≤ x ≤ 150.
A′ (x) = −4x + 300
which has root x = 75, so that by the Extreme Value Theorem we only need to check
x = 0, 75, 150
A(0) = 0
A(75) = 75(300 − 150) = 75(150) > 0
A(150) = 0
so the area is maximized when x = 75 ft. and y = (300 − 150)/3 = 150/3 = 50 ft.
16
Problem 17
As the proprietor of “With Pipe and Math Book” located near Great Swamp, New
Jersey off I-287, you annually sell 1000 corncob pipes. Each pipe costs you 0.20$ wholesale and 2$ a year to store. Each shipment costs you 10$. Assuming that the pipes are
sold at a constant rate and that each new shipment arrives exactly when your inventory
is depleted, how many pipes should you order at one time in order to minimize your costs?
Hint: your answer should be a whole number. What might you do if you had not gotten
a whole number answer?
What are the competing factors? What factor(s) do not affect the result?
The total cost C can be given in terms of the number of pipes in each shipment x. We
have
total cost = storage cost + shipment cost + cost of pipes
so that, since the average number of pipes on a shelf is x/2 and the number of shipments
is 1000/x, therefore
x
1000
C(x) = 2 + 10
+ 1000(0.2)
2
x
10, 000
+ 200
=x+
x
= x + 10, 000x−1 + 200
and
C ′ (x) = 1 − 10, 000x−2
so that the critical values occur where
10, 000
=1
x2
and solving for x yields
x = ±100
So that the only value which makes sense for the answer is
x = 100.
If you had not gotten a whole number answer, then it is a good idea to test the nearest
two whole numbers and take the bigger one of them (although this is not in and of itself
mathematically guaranteed to yield the best answer, but it’s still a good idea)
17
Problem 18
Evaluate the following integrals:
a.
Z
x2 ex dx
u = x2
du = 2xdx
dv = ex dx
v = ex
so
Z
2 x
2 x
x e dx = x e −
Z
2xex dx
u = 2x
du = 2dx
dv = ex dx v = ex
so
Z
2 x
2 x
x e dx = x e −
Z
Z
x
x
2xe dx = x e − 2xe − 2e dx = x2 ex − 2xex + 2ex .
x
2 x
b.
Z
x ln(x)dx
u = ln(x) du = x1 dx
2
dv = xdx v = x2
so that
Z
Z 2
x 1
x2
dx
x ln(x)dx = ln(x) −
2
2 x
Z
x
x2
dx
= ln(x) −
2
2
x2 x2
= ln(x) −
2
4
18
c.
Z
2
xex dx
1
u=x
du = dx
x
dv = e dx v = ex
so that
Z
2
x
Z
2
ex dx
1
2
= 2e − e − ex |21
xe dx =
1
xex |21
−
= 2e2 − e − (e2 − e)
= e2
d.
Z
2
xex
2 +1
dx
1
Use a U-substitution. Let
u = x2 + 1
then
du = 2xdx.
also changing the limits of integration gives
Z
Z 2
x2 +1
xe
dx =
1
2
5
1 u
e du
2
1
= eu |52
2
1
1
= e5 − e2
2
2
19
Estimating integrals
R2
Estimate 0 sin(x)dx using a Riemann sum (regular right endpoints) and the trapezoid
rule, with 4 intervals. Draw the graph indicating the Riemann sum (but not the trapezoid
rule)
So using a Riemann sum we get
Z 2
1
sin(x)dx ≃ (sin(3/2) + sin(2) + sin(5/2) + sin(3))
2
0
and using the trapezoid rule we get
Z 2
1 sin(1) + sin(3/2) sin(3/2) + sin(2) sin(2) + sin(5/2) sin(5/2) + sin(3)
+
+
+
sin(x)dx ≃
2
2
2
2
2
0
20
Taylor series
Estimate e2 using a third order Taylor polynomial.
x2 x3
+
e ≃1+x+
2!
3!
x
so
e2 ≃ 1 + 2 +
22 23
+
2!
3!
21
2nd fundamental theorem
Find the derivatives of the following:
f (x) =
Z
x
arctan(t)dt
3
By the second fundamental theorem you get:
f ′ (x) = arctan(x)
g(x) =
Z
sin x
t2 dt
7
by the second fundamental theorem and the chain rule we get
g ′ (x) = (sin(x))2 cos(x)
22
Average value
Find the average height of f (x) = sin(x) on the interval [0, π]. It is
Rπ
sin(x)dx
1
0
= (− cos(x)|π0 )
π−0
π
1
= (−(−1) − (−1))
π
2
=
π
The average slope over the interval is of course given by the slope of the secant line
(chord) and that is zero in this case.