Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Solutions to Math 1131 Test #1: Limits, Continuity, and Differentiation, Exams of Analytical Geometry and Calculus

Solutions to a math test covering limits, continuity, and differentiation. It includes numerical and analytical solutions for finding limits, determining points of discontinuity, finding values of constants that make a function continuous, and calculating derivatives using the definition of a derivative. It also covers the equation of the tangent line and the use of the intermediate value theorem.

Typology: Exams

Pre 2010

Uploaded on 08/04/2009

koofers-user-onw-1
koofers-user-onw-1 🇺🇸

0

(1)

10 documents

1 / 6

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Test #1, September 11th, 2007, Solutions
Math 1131
1.Determine the following limits numerically and analytically:
(a) lim
x→∞ µ2x+ 1
2x33x
(b) lim
x0
1cos 5x
x2(c) lim
x1
5
x1
3
x1
Solution: (a) Let f(x) = ¡2x+1
2x3¢3xfor x > 2. Some of the values of fat inputs that are
getting larger and larger are tabulated next:
f(10) 566.384292607228022196978071097
f(100) 415.862133853092606249646578271
f(1000) 404.642314884456573360577354755
f(10000) 403.549854412652369870990664109
f(100000) 403.440896679290503911843930292
f(1000000) 403.430003782343038726003184243
It seems to be the case that the limit is 403.4. To do this algebraically we use the first
fundamental limit:
lim
x→∞ µ2x+ 1
2x33x
= lim
x→∞ µ1 + 4
2x33x
= lim
x→∞ "µ1 + 1
(2x3)/42x3
4#(3x)4
2x3
=e
lim
x→∞
12
23/x =e6
so
lim
x→∞ µ2x+ 1
2x33x
=e6403.42879349... .
¥
(b) If g(x) = 1cos 5x
x2if x6= 0. Some of the values of gfor inputs getting closer and
closer to zero are included in the next table:
g(-0.1) 12.2417438109627283883718417396
g(0.01) 12.4973960503375343712918884300
g(-0.001) 12.4999739583550347125341050000
g(0.0001) 12.4999997395833355034722000000
g(-0.00001) 12.4999999973958333335500000000
g(0.000001) 12.4999999999739583330000000000
We can guess that this limit must be equal to 12.5 or 25
2. This is indeed the cases since
lim
x0
1cos 5x
x2= lim
x0
2 sin2(5x
2)
x2= 2 lim
x0µsin 5x
2
5x
22µ5
22
= 225
4=25
2.
Hence lim
x0
1cos 5x
x2=25
2.¥
I
pf3
pf4
pf5

Partial preview of the text

Download Solutions to Math 1131 Test #1: Limits, Continuity, and Differentiation and more Exams Analytical Geometry and Calculus in PDF only on Docsity!

Test # 1, September 11 th, 2007, Solutions Math 1131

  1. Determine the following limits numerically and analytically:

(a) lim x→∞

2 x + 1 2 x − 3

) 3 x (b) lim x→ 0

1 − cos 5x x^2

(c) lim x→ 1

√ (^5) x − 1 √ (^3) x − 1

Solution: (a) Let f (x) =

( 2 x+ 2 x− 3

) 3 x for x > 2. Some of the values of f at inputs that are getting larger and larger are tabulated next: f(10) 566. f(100) 415. f(1000) 404. f(10000) 403. f(100000) 403. f(1000000) 403. It seems to be the case that the limit is ≈ 403 .4. To do this algebraically we use the first fundamental limit:

lim x→∞

2 x + 1 2 x − 3

) 3 x = lim x→∞

2 x − 3

) 3 x = lim x→∞

[(

(2x − 3)/ 4

) 2 x 4 − 3 ](3 2 xx−)4 3 = e

lim x→∞

2 − 3 /x (^) = e^6

so

lim x→∞

2 x + 1 2 x − 3

) 3 x = e^6 ≈ 403. 42879349 ....

(b) If g(x) =

1 − cos 5x x^2

if x 6 = 0. Some of the values of g for inputs getting closer and

closer to zero are included in the next table: g(-0.1) 12. g(0.01) 12. g(-0.001) 12. g(0.0001) 12. g(-0.00001) 12. g(0.000001) 12. We can guess that this limit must be equal to 12.5 or 252. This is indeed the cases since

lim x→ 0

1 − cos 5x x^2

= lim x→ 0

2 sin^2 (^52 x ) x^2

= 2 lim x→ 0

sin 52 x 5 x 2

Hence lim x→ 0

1 − cos 5x x^2

I

(c) Finally if h(x) =

√ (^5) x − 1 √ (^3) x − 1 for every real number x. Some of the values of g around

zero are shown below: h(0.9) 0. h(1.01) 0. h(0.999) 0. h(1.0001) 0. h(0.99999) 0. h(1.000001) 0. So it is reasonable to conclude that the limit of this function at x = 1 is 0.6 = 3/5. This is true since if we make the change of variable x = (1 + t)^3 we see that t → 0 and

lim x→ 1

√ (^5) x − 1 √ (^3) x − 1 = lim t→ 0

(1 + t) (^35) − 1 t

Therefore, lim x→ 1

√ (^5) x − 1 √ (^3) x − 1 = 3/ 5. •

  1. Determine if the following functions are continuous or not. If they are not continuous find the points of discontinuity.

(a) f (x) =

x^2 + x if x ≥ 0 sin 3x x

if x < 0

, (b) g(x) =

ln(^1 e + x) if x ≥ 0 ln(1 + |x|) x

if x < 0

Solution: (a) The function f is continuous at every point except maybe x = 0. At x = 0

we have lim x→ 0 +^

x^2 + x = 0 = f (0). and lim x→ 0 −^

f (x) = lim x→ 0 −

sin 3x x

= 3 lim x→ 0 −

sin 3x 3 x

= 3. Therefore

the function f is discontinuous at x = 0 but continuous at every other point. (b) The function g is continuous at every point except maybe x = 0. The limit from the right hand side of 0 is lim x→ 0 +^

ln(1/e + x) = ln 1/e = −1 = g(0). For the limit from the left at

0 we get

lim x→ 0 −^

g(x) = lim x→ 0 −

ln(1 + |x|) x

= lim x→ 0 −

ln(1 + |x|) |x|

|x| x

So g is continuous. •

  1. Find all values of a such that the following function is continuous:

h(x) =

ax 2 + a^2 x

if x ≥ 1

√ (^3) x − 1

x − 1

if x < 1

II

which gives h′(x) =

2 x + 3

, x 6 = −

(b) Similarly

i′(a) = lim x→a

i(x) − i(a) x − a

= lim x→a

x+ 2 x− 3 −^

a+ 2 a− 3 x − a

= lim x→a

(x + 1)(2a − 3) − (a + 1)(2x − 3) (x − a)(2x − 3)(2a − 3)

lim x→a

(2x − 3)(2a − 3)

(2a − 3)^2

which gives i′(x) = −

(2x − 3)^2

(c, Bonus:) We have

k′(a) = lim x→a

k(x) − k(a) x − a

= lim x→a

x^2 x− 1 −^

a^2 a− 1 x − a

= lim x→a

x^2 (a − 1) − a^2 (x − 1) (x − a)(x − 1)(a − 1)

lim x→a

ax(x − a) − (x − a)(x + a) (x − a)(x − 1)(a − 1)

= lim x→a

ax − x − a (x − 1)(a − 1)

a^2 − 2 a (a − 1)^2

and so k′(x) =

x(x − 2) (x − 1)^2

, x 6 = 1..

  1. For the functions in Problem 5 find the equation of the tangent line at x = 3 and plot both the tangent line and the corresponding function.

Solution: (a) The equation of the tangent line is y = h(3) + h′(3)(x − 3). Since h(3) = 3

and h′(1) = 1/3 we get y = 3 + (x − 3)/3 or y = (x + 6)/3. The graph of h and of the

tangent line is included below:

4

3

x

2

1

(^00246)

(b) Since i(3) = 4/3 and i′(3) = −^59 the equation of the tangent line at (3, 4 /3) is

y =

27 − 5 x 9

. The graph is

IV

x

2

3

4

1

2.5 3.5 5

2 3

(c) In this case k(3) = 9/2 and k′(3) = 3/4. Then the equation of tangent line becomes

y = 9/2 + 3(x − 4)/4 or y = (3x + 9)/ 4 and the graph:

4.5 5

6

3

x

4

5

2 2.

4

  1. Use the Intermediate Value Theorem to show that the following equation has a solution in the specified interval:

tan 2x = x in (

π 4

3 π 4

Solution: Consider the function f (x) = sin 2x−x cos 2x defined for all real x. All elementary functions involved are defined for all real numbers. Hence f is continuous on [π 4 , 34 π ]. Since f (π 4 ) = sin π 2 − π 4 cos π 2 = 1 and f (^34 π ) = sin 32 π − 34 π cos 32 π = −1 we see that 0 is between f (π 4 ) and f (^34 π ). Hence by IVT we know there is a number x ∈ (π 4 , 34 π ) such that f (x) = 0. This means sin 2x − x cos 2x = 0 or tan 2x = x. •

  1. A particle moves along a straight line with equation of motion s(t) = 4t^3 − 15 t^2 + 12 t + 7, where s is measured in meters and t in seconds. Find the time when the velocity becomes zero.

Solution: We calculate the speed v := dsdt = 12t^2 − 30 t+12 = 6(2t^2 − 5 t+2) = 6(2t−1)(t−2). Therefore, v(t) = 0 implies t = 0.5 seconds and t = 2 seconds. •

  1. Let f (x) = | sin(x)| defined for all real numbers x. Where is f differentiable? What about g(x) = | sin(x)|^3?

V