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Spring 2008 Math 116 Exam III Review, Exams of Algebra

Solutions to various math problems covered in the spring 2008 math 116 exam, including partial fraction decomposition, inverse functions, limits, and quadratic equations.

Typology: Exams

Pre 2010

Uploaded on 09/02/2009

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Spring 2008
04/24/2008
Math 116 - Exam III Review
1. (A2) Find the partial fraction decomposition of 9x+ 21
(x+ 5)(xโˆ’3).
Solution:
There are two 1-repeating linear factors. We thus have
9x+ 21
(x+ 5)(xโˆ’3) =A
x+ 5 +B
xโˆ’3=A(xโˆ’3) + B(x+ 5)
(x+ 5)(xโˆ’3) =(A+B)xโˆ’3A+ 5B
(x+ 5)(xโˆ’3)
We now need to solve a system of equations:
A+B= 9
โˆ’3A+ 5B= 21
3A+ 3B= 27
โˆ’3A+ 5B= 21
8B= 48
B= 6
A= 3
Thus, 9x+ 21
(x+ 5)(xโˆ’3) =3
x+ 5 +6
xโˆ’3.
2. (3.6) Find the inverse of f(x) = โˆšxโˆ’3. Then show that the function you found is
indeed the inverse.
Solution:
Note, that fis one-to-one (can be seen from a picture). Now find the domain of f. It
is {xโˆˆR|xโ‰ฅ3}= [3,โˆž). Find the inverse by solving y=โˆšxโˆ’3 for x:
y=โˆšxโˆ’3
y2=xโˆ’3
y2+ 3 = x
fโˆ’1(x) = x2+ 3
We finally need to specify the domain of fโˆ’1(x). It is defined to be the range of f.
Therefore,
domain of fโˆ’1= range of f={xโˆˆR|xโ‰ฅ0}= [0,โˆž).
To show that the function fโˆ’1is indeed the inverse of f, we need to check the following
two conditions:
(fโ—ฆfโˆ’1)(x) = xโˆ€xโˆˆdomain(fโˆ’1)
(fโˆ’1โ—ฆf)(x) = xโˆ€xโˆˆdomain(f).
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Spring 2008 04/24/

Math 116 - Exam III Review

  1. (A2) Find the partial fraction decomposition of

9 x + 21 (x + 5)(x โˆ’ 3)

Solution: There are two 1-repeating linear factors. We thus have

9 x + 21 (x + 5)(x โˆ’ 3)

A

x + 5

B

x โˆ’ 3

A(x โˆ’ 3) + B(x + 5) (x + 5)(x โˆ’ 3)

(A + B)x โˆ’ 3 A + 5B (x + 5)(x โˆ’ 3)

We now need to solve a system of equations:

A + B = 9 โˆ’ 3 A + 5B = 21

3 A + 3B = 27

โˆ’ 3 A + 5B = 21

8 B = 48

B = 6

A = 3

Thus, 9 x + 21 (x + 5)(x โˆ’ 3)

x + 5

x โˆ’ 3

  1. (3.6) Find the inverse of f (x) =

x โˆ’ 3. Then show that the function you found is indeed the inverse. Solution: Note, that f is one-to-one (can be seen from a picture). Now find the domain of f. It is {x โˆˆ R| x โ‰ฅ 3 } = [3, โˆž). Find the inverse by solving y =

x โˆ’ 3 for x:

y =

x โˆ’ 3 y^2 = x โˆ’ 3 y^2 + 3 = x f โˆ’^1 (x) = x^2 + 3

We finally need to specify the domain of f โˆ’^1 (x). It is defined to be the range of f. Therefore, domain of f โˆ’^1 = range of f = {x โˆˆ R| x โ‰ฅ 0 } = [0, โˆž). To show that the function f โˆ’^1 is indeed the inverse of f , we need to check the following two conditions:

(f โ—ฆ f โˆ’^1 )(x) = x โˆ€x โˆˆ domain(f โˆ’^1 ) (f โˆ’^1 โ—ฆ f )(x) = x โˆ€x โˆˆ domain(f ).

We have

(f โ—ฆ f โˆ’^1 )(x) = f (f โˆ’^1 (x)) = f (x^2 + 3) =

x^2 + 3 โˆ’ 3 =

x^2 = |x| = x (f โˆ’^1 โ—ฆ f )(x) = f โˆ’^1 (f (x)) = f โˆ’^1 (

x โˆ’ 3) = (

x โˆ’ 3)^2 + 3 = x โˆ’ 3 + 3 = x.

Thus, the function f โˆ’^1 is really the inverse of f.

  1. (5.1)

(a) Define the Euler-Mascheroni constant e. (b) A promissory note will pay $30, 000 at maturity 10 years from now. How much would you pay for the note now if the note gains value at a rate of 4% compounded continuously? What would be the noteโ€™s value today if it gained value at a rate of 4% compounded monthly?

Solution:

(a) limxโ†’โˆž

1 + (^1) x

)x

(b) The continuous compound interest formula is given by

A = P ert.

We are interested in the value of the note today, i.e. P. Solving for P we obtain

P =

A

ert^

Now setting A = $30, 000, r = 0.04, t = 10, we get

P =

e^0.^04 ยท^10

If the interest is compounded monthly, we use the formula

Pt = P 0

r m

)mt ,

where m is the number of compounding periods per year and t is the time in years. Solving for P 0 we get

P 0 = Pt

r m

)โˆ’mt .

Letting t = 10, P 10 = $30, 000, and r = 0.04, we obtain

P 0 = $30, 000

  1. (5.4) The magnitude of an earthquake is commonly measured using the Richter scale given by M = 23 log (^) EE 0 , where E is the energy released by the earthquake and E 0 = 104.^40. To inflict serious damage, an earthquake requires a magnitude of over 5.6 on the Richter scale. How many times more powerful than this was the great 1906 Colombia earthquake, which registered a magnitude of 8.6 on the Richter scale?

From the first row we see immediately, that c = 24. Thus

102 a + 10b + 24 = 30 202 a + 20b + 24 = 34

or

102 a + 10b = 6 202 a + 20b = 10

2 ยท 102 a + 20b = 12 202 a + 20b = 10

(200 โˆ’ 400)a = 2 โˆ’ 0 .01 = a

Solving for b we obtain b = 101 (6 โˆ’ 100 ยท (โˆ’ 0 .01)) = 0.7. Thus,

f (x) = โˆ’ 0. 01 x^2 + 0. 7 x + 24,

and f (30) = โˆ’ 0. 01 ยท 302 + 0. 7 ยท 30 + 24 = 36.

  1. (10.3) Consider the augmented matrix ๏ฃซ ๏ฃฌ ๏ฃฌ ๏ฃญ

(a) Using the variables x, y, and z, what system does this augmented matrix repre- sent? (b) Either solve the system of equations or argue why it has no solution.

Solution:

(a) 2 x + 2y + 1z = 7 1 x + 3y + 3z = 8 1 z = 5 2 z = 6

(b) The system has no solution: ๏ฃซ ๏ฃฌ ๏ฃฌ ๏ฃญ

This is a contradiction.

  1. (10.3+10.4) Consider the system of equations

x + 2y = 7 2 x + 3y = 6.

(a) Solve this system of equations by Gauss-Jordan elimination, and check your an- swer. (b) Solve this system of equations by writing it in the form AX = B. Find the inverse of A and use it to solve the equation AX = B. Show your calculations.

Solution:

(a) ( 1 2 7 2 3 6

(2) 7 โ†’ 2 ยท(1)โˆ’(2) โˆ’โˆ’โˆ’โˆ’โˆ’โˆ’โˆ’โˆ’โ†’

(1) 7 โ†’(1)โˆ’ 2 ยท(2) โˆ’โˆ’โˆ’โˆ’โˆ’โˆ’โˆ’โˆ’โ†’

Now check your answer:

โˆ’9 + 2 ยท 8 = 7X 2 ยท (โˆ’9) + 3 ยท 8 = 6X

(b) (^) ( 1 2 2 3

x y

Now find the inverse of A =

(2) 7 โ†’ 2 ยท(1)โˆ’(2) โˆ’โˆ’โˆ’โˆ’โˆ’โˆ’โˆ’โˆ’โ†’

(1) 7 โ†’(1)โˆ’ 2 ยท(2) โˆ’โˆ’โˆ’โˆ’โˆ’โˆ’โˆ’โˆ’โ†’

Thus, Aโˆ’^1 =

and so

( x y

  1. Decide whether or not the following statements are true.

(a) If A is an m ร— n-, and B an n ร— m-matrix, then AB is defined and is an n ร— n- matrix. (b) Every non-singular matrix has a multiplicative inverse. (c) If A and B are square matrices with the same dimension, then AB = BA. (d) If both A and B are m ร— n-matrices, then A + B = B + A. (e) Every system of linear equations can be written in the form AX = B and has the solution X = Aโˆ’^1 B.