Review Problems

(largely taken from old midterms)

Problem 1

Find the following limits or find that the limit does not exist (you may NOT usel’Hopital’s rule)

a.

limx→0

tan(5x)

x= lim

x→05sin(5x)

5x

1

cos(5x)

= 5 · 1 · 1

1= 5

b.

limx→2

|x − 2|x − 2

Calculate the one sided limits separately:

limx→2+

|x − 2|x − 2

= limx→2+

x − 2

x − 2= 1

and

limx→2−

|x − 2|x − 2

= limx→2−

−(x − 2)

x − 2= −1

1

2

Problem 2

Find A and B such that the following function is continuous

f(x) =

x2 x ≤ 0Ax + B 0 < x ≤ 25√

x−5x−25

25 < x

Use the definition of continuity applied at x = 0 and x = 25. In other words find fourone-sided limits and equate the appropriate pairs to get equations for A and B:

limx→0−

f(x) = limx→0−

x2 = 0

limx→0+

f(x) = limx→0+

Ax + B = B

limx→25−

f(x) = limx→25−

Ax + B = 25A + B

limx→25+

f(x) = limx→25+

√x − 5

x − 25

√x + 5√x + 5

= limx→25+

x − 25

x − 25

1√x + 5

=1

10

And so you get the two equations

0 = B

25A + B =1

10

which give A = 1250

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 thenumber and location of solutions to the equation f(x) = 0? State any theorems that youare using.

The following picture indicates the answer, that there are at least three solutions, atx = −1, x = 5 and one in the interval (6, 8) according to the Intermediate Value Theorem.

Problem 4

Find the derivative of f(x) = 1x

USING ONLY THE DEFINITION OF THE DERIV-ATIVE.

f ′(x) ≡ limh→0

f(x + h) − f(x)

h= lim

h→0

1x+h

− 1x

h

(x + h)(x)

(x + h)(x)

= limh→0

x − (x + h)

h(x + h)(x)

= limh→0

−h

h(x + h)(x)

= limh→0

−1

(x + h)(x)

=−1

x2

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)exx4.2 + 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

y

dy

dx= sin(x) + x cos(x).

Then solving for dydx

:

dy

dx= y(sin(x) + x cos(x))

and substituting for y:

dy

dx= xsin(x)(sin(x) + x cos(x))

5

Problem 6

Find the tangent line to

3x3 + 2y2 − 2x + 8y = 20

through the point (2,−4).Use implicit differentiation:

9x2 + 4ydy

dx− 2 + 8

dy

dx= 0

and solving for dydx

:

dy

dx(4y + 8) = 2 − 9x2 ⇒ dy

dx=

2 − 9x2

4y + 8

so the slope of the tangent line is

2 − 9 · 22

4 · −4 + 8=

2 − 36

−16 + 8=

−34

−8=

17

4.

then using the slope form of the line we get that the tangent is:

17

4=

y + 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 areusing and why it works.

Relative Maximum: x = −2 either 1st (p′(x) switches from positive to negative aroundx = −2) or 2nd (p′′(−2) < 0) derivative tests

Relative Minimum: x = 5 either 1st (p′(x) switches from negative to positive aroundx = 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

h(x) =(x − 3)(x − 4)(x − 1)

(x + 1)(x − 2)(x + 4)

There are vertical asymptotes at x = −1, 2,−4.

h(x) =(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

x3 + 3x2 − 6x − 8

So

limx→±∞

h(x) = limx→±∞

x3 − 8x2 + 19x − 12

x3 + 3x2 − 6x − 8

(

1/x3

1/x3

)

= limx→±∞

1 − 8/x + 19/x2 − 12/x3

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 orneither:

f(x) = x4

5 (x − 1)

g(x) = x3

5 (x − 1)

f(x) = x9/5 − x4/5

so

f ′(x) =9

5x4/5 − 4

5x−1/5

=x−1/5

5(9x − 4)

=1

5x1/5(9x − 4)

and so f ′(x) has an asymptote at x = 0 where f(x) is defined. Further

limx→0+

f ′(x) = −∞

limx→0−

f ′(x) = +∞

and since the signs are opposite f(x) has a cusp.

g(x) = x8/5 − x3/5

so

g′(x) =8

5x3/5 − 3

5x−2/5

=x−2/5

5(8x − 3)

=1

5x2/5(8x − 3)

and so g′(x) has an asymptote at x = 0 where g(x) is defined. Further

limx→0+

g′(x) = −∞

limx→0−

g′(x) = −∞

and since the signs are the same g(x) has a vertical tangent.

10

Problem 11

Use a linear approximation to estimate3√

8.1

Let

f(x) = 3

√

(x) = x1/3

so

f ′(x) =1

3x−2/3

and the linear approximation to f(x) at x = 8 is

L(x) = f(8) + f ′(8)(x − 8)

= 3

√

(8) +1

3(8)−2/3(x − 8)

= 2 +1

3 · 82/3(x − 8)

= 2 +1

12(x − 8)

and then

3√

8.1 ≃ L(8.1) = 2 +1

12(8.1 − 8) = 2 +

0.1

12= 2 +

1

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 NOTneed to simplify the second iterate).

Recall the iterative formula

xn+1 = xn − f(xn)

f ′(xn)

and also

q′(x) = 3x2 − 4x − 5.

We begin with x0 = −1.

q(−1) = −1 − 2 + 5 + 6 = 8

q′(−1) = 3 + 4 − 5 = 2.

so

x1 = −1 − 8

2= −5

And continuing one more step

q(−5) = −125 − 2(25) + 25 + 6 = −144

q′(−5) = 75 + 20 − 5 = 90

so

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 circumferenceis 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

dc

c100% =

0.2

5100% =

20

5% = 4%.

b) Recalling the formula

V =1

4πc2h

for the volume of a cylinder in terms of its circumference c and height h, find the calcu-lated volume of the cylinder.

V =1

4π(5)2(4) =

25

πin3

c) Find the percentage error in the calculated volume.We have

dV =2h

4πcdc

so that

dV

V=

2h4π

cdc14π

c2h=

2dc

c

and thus∆V

V100% ≃ dV

V100 = 2

dc

c100 = 2(4) = 8%.

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 betweent = 0 and t = 4.

displacement = s(4) − s(0) = 5 − 1 = 4.

ds(t)

dt= v(t) = 3t2 − 12t + 9

= 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

s(4) = 64 − 6(16) + 36 + 1 = 101 − 96 = 5.

Therefore

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 movingmore and more slowly away from you horizontally with speed e−t2 ft/s and it is rising at3 ft/s. Assume that the kite starts on the ground some distance away from you and thatafter 10 seconds, the kite is 40 ft distant from you horizontally. Find out how fast thestring is unwinding at that moment.

Given:

dx

dt= e−t2 dy

dt= 3x(10) = 40

We have

l2 = x2 + y2

so that

2ldl

dt= 2x

dx

dt+ 2y

dy

dtso that

dl

dt=

xdxdt

+ y dydt

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

dt|t=10 =

40e−100 + 30 · 350

=

(

4

5e−100 +

9

5

)

ft/s

15

Problem 16

You are building a fence in the partitioned rectangular shape as pictured below. Thetotal 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 checkx = 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, NewJersey off I-287, you annually sell 1000 corncob pipes. Each pipe costs you 0.20$ whole-sale and 2$ a year to store. Each shipment costs you 10$. Assuming that the pipes aresold at a constant rate and that each new shipment arrives exactly when your inventoryis 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 gottena 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 shipmentsis 1000/x, therefore

C(x) = 2x

2+ 10

1000

x+ 1000(0.2)

= x +10, 000

x+ 200

= x + 10, 000x−1 + 200

and

C ′(x) = 1 − 10, 000x−2

so that the critical values occur where10, 000

x2= 1

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 nearesttwo whole numbers and take the bigger one of them (although this is not in and of itselfmathematically guaranteed to yield the best answer, but it’s still a good idea)

17

Problem 18

Evaluate the following integrals:

a.∫

x2exdx

u = x2 du = 2xdxdv = exdx v = ex

so∫

x2exdx = x2ex −∫

2xexdx

u = 2x du = 2dxdv = exdx v = ex

so∫

x2exdx = x2ex −∫

2xexdx = x2ex −(

2xex −∫

2exdx

)

= x2ex − 2xex + 2ex.

b.∫

x ln(x)dx

u = ln(x) du = 1xdx

dv = xdx v = x2

2

so that∫

x ln(x)dx = ln(x)x2

2−

∫

x2

2

1

xdx

= ln(x)x2

2−

∫

x

2dx

= ln(x)x2

2− x2

4

18

c.∫ 2

1

xexdx

u = x du = dxdv = exdx v = ex

so that∫ 2

1

xexdx = xex|21 −∫ 2

1

exdx

= 2e2 − e −(

ex|21)

= 2e2 − e − (e2 − e)

= e2

d.∫ 2

1

xex2+1dx

Use a U-substitution. Let

u = x2 + 1

then

du = 2xdx.

also changing the limits of integration gives∫ 2

1

xex2+1dx =

∫ 5

2

1

2eudu

=1

2eu|52

=1

2e5 − 1

2e2

19

Estimating integrals

Estimate∫ 2

0sin(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 trapezoidrule)

So using a Riemann sum we get∫ 2

0

sin(x)dx ≃ 1

2(sin(3/2) + sin(2) + sin(5/2) + sin(3))

and using the trapezoid rule we get∫ 2

0

sin(x)dx ≃ 1

2

(

sin(1) + sin(3/2)

2+

sin(3/2) + sin(2)

2+

sin(2) + sin(5/2)

2+

sin(5/2) + sin(3)

2

)

20

Taylor series

Estimate e2 using a third order Taylor polynomial.

ex ≃ 1 + x +x2

2!+

x3

3!so

e2 ≃ 1 + 2 +22

2!+

23

3!

21

2nd fundamental theorem

Find the derivatives of the following:

f(x) =

∫ x

3

arctan(t)dt

By the second fundamental theorem you get:

f ′(x) = arctan(x)

g(x) =

∫ sin x

7

t2dt

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∫ π

0sin(x)dx

π − 0=

1

π(− cos(x)|π0 )

=1

π(−(−1) − (−1))

=2

πThe average slope over the interval is of course given by the slope of the secant line

(chord) and that is zero in this case.