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Solutions to various math problems covered in the spring 2008 math 116 exam, including partial fraction decomposition, inverse functions, limits, and quadratic equations.
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Spring 2008 04/24/
9 x + 21 (x + 5)(x โ 3)
Solution: There are two 1-repeating linear factors. We thus have
9 x + 21 (x + 5)(x โ 3)
x + 5
x โ 3
A(x โ 3) + B(x + 5) (x + 5)(x โ 3)
(A + B)x โ 3 A + 5B (x + 5)(x โ 3)
We now need to solve a system of equations:
A + B = 9 โ 3 A + 5B = 21
Thus, 9 x + 21 (x + 5)(x โ 3)
x + 5
x โ 3
x โ 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 โ R| x โฅ 3 } = [3, โ). Find the inverse by solving y =
x โ 3 for x:
y =
x โ 3 y^2 = x โ 3 y^2 + 3 = x f โ^1 (x) = x^2 + 3
We finally need to specify the domain of f โ^1 (x). It is defined to be the range of f. Therefore, domain of f โ^1 = range of f = {x โ R| x โฅ 0 } = [0, โ). To show that the function f โ^1 is indeed the inverse of f , we need to check the following two conditions:
(f โฆ f โ^1 )(x) = x โx โ domain(f โ^1 ) (f โ^1 โฆ f )(x) = x โx โ domain(f ).
We have
(f โฆ f โ^1 )(x) = f (f โ^1 (x)) = f (x^2 + 3) =
x^2 + 3 โ 3 =
x^2 = |x| = x (f โ^1 โฆ f )(x) = f โ^1 (f (x)) = f โ^1 (
x โ 3) = (
x โ 3)^2 + 3 = x โ 3 + 3 = x.
Thus, the function f โ^1 is really the inverse of f.
(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 noteโs value today if it gained value at a rate of 4% compounded monthly?
Solution:
(a) limxโโ
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
ert^
Now setting A = $30, 000, r = 0.04, t = 10, we get
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
)โmt .
Letting t = 10, P 10 = $30, 000, and r = 0.04, we obtain
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 โ 400)a = 2 โ 0 .01 = a
Solving for b we obtain b = 101 (6 โ 100 ยท (โ 0 .01)) = 0.7. Thus,
f (x) = โ 0. 01 x^2 + 0. 7 x + 24,
and f (30) = โ 0. 01 ยท 302 + 0. 7 ยท 30 + 24 = 36.
(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.
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)โ(2) โโโโโโโโโ
(1) 7 โ(1)โ 2 ยท(2) โโโโโโโโโ
Now check your answer:
โ9 + 2 ยท 8 = 7X 2 ยท (โ9) + 3 ยท 8 = 6X
(b) (^) ( 1 2 2 3
x y
Now find the inverse of A =
(2) 7 โ 2 ยท(1)โ(2) โโโโโโโโโ
(1) 7 โ(1)โ 2 ยท(2) โโโโโโโโโ
Thus, Aโ^1 =
and so
( x y
(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 = Aโ^1 B.