Factorize

Problem:

Factorize $ 6x^2 - 7y^2 $

Solution:

$ (\sqrt{6}x + \sqrt{7}y)(\sqrt{6}x - \sqrt{7}y) $

Interval Notation (5)

Problem:



Solution:

$ C \cup D = (-\infty, \infty) $
$ C \cap D = [2,9) $

Interval Notation (4)

Problem:



Solution:

$ F \cap H = [5, \infty) $
$ F \cup H = (4, \infty) $

Interval Notation (3)

Problem:



Solution:

$ A \cup B = (-\infty, 1] \cup (6, \infty) $
$ A \cap B = \emptyset $

Interval Notation (2)

Problem:



Solution:

$ E \cap F  = (4, 5] $
$ E \cup F  = (-\infty, +\infty) $

Interval notation

Problem:



Solution:

$ A \cup B = (-\infty, 8] $
$ A \cap B = (-\infty, 3] $

Logarithm

Problem:




Solution:

$ \begin{eqnarray*} & & 6\log(25-x^2)-(\log(5+x)+\log(5-x)) \\ &=& 6\log(25-x^2)-(\log(5+x)(5-x)) \\ &=& 6\log(25-x^2)-\log(25-x^2) \\ &=& 5\log(25-x^2) \\ &=& \log(25-x^2)^5 \\ \end{eqnarray*} $

Rate of change

Problem:

¨

Solution:

a) -1
b) -6
c) 10.546875
d) 0.127706406

Angles

Problem:

Solution:

1) 180
2) PQS
3) 130
4) 30
5) 35
6) 100
7) 50
8) 130
9) x = 11, m<DEF = 29, m<FEG = 61
10) x = 10, m <DEF = 91, m<FEG = 89
11) 2, 4
12) 2, 3
13) Right
14) m<KLM = m<KLN = 45

Matrix

Problem:



Solution:

$ \left(\begin{array}{cc} 4 & -5 \\ -2 & 3\end{array}\right) + \left(\begin{array}{cc} 7 & -6 \\ 1 & -4\end{array}\right) = \left(\begin{array}{cc} 11 & -11 \\ -1 & -1\end{array}\right) $

Summation

Problem:



Solution:

$ \bar{x} = 10.8 $.

Here are the values of absolute difference from the mean

6.8

4.8

0.8

3.2

9.2

Here are the weighted absolute difference from the mean

27.2

14.4

1.6

3.2

9.2

The sum of these numbers is 55.6

Therefore the final result is 5.05454545...

Extra credits

Problem:



Solution:

$ f'(x) = 3ax^2 + 2bx + c $
$ f''(x) = 6ax + 2b $

The point of inflection has x coordinate -1, therefore

$ 6a(-1) + 2b = 0 $

The extremas has x coordinate 1 and -3, therefore, 1 and -3 are the roots of  $3ax^2 + 2bx + c = 0 $

$ 3a + 2b + c = 0 $
$ 3a(-3)^2 + 2b(-3) + c = 0 $.

The three equations tell us $ b = 3a, c = -9a $.

So the equation is $ f(x) = ax^3 + 3ax^2 - 9ax + d$

When x = 1, $ f(x) = a + 3a - 9a + d = -5a + d = -9 $
When x = -1, $ f(x) = -a + 3a + 9a + d = 11a + d = 23 $

Therefore $ a = 2, d = 1 $

The final answer is $ 2x^3 + 6x^2 - 18x + 1 $

Point of inflection

Problem:



Solution:

The point of inflection is where the second derivative is 0, so we compute the second derivative as follow:

$ f'(x) = -3x^2 + 18x - 4 $
$ f''(x) = -6x + 18 $

Therefore the point of inflection has x coordinate 3, the y coordinate is $ -(3)^3 + 9 \times 3^2 - 4 \times 3 - 1 = 119 $

The curve is concave up in $ (-\infty, 3) $ and is concave down in $ (3, \infty) $.

Solid

Problem:



Solution:

Minimize $ 4\pi r^2 + 2\pi r h $ subject to $ \frac{4}{3}\pi r^3 + \pi r^2 h = 6 $.

The Lagrangian is $ 4\pi r^2 + 2\pi r h - \lambda(\frac{4}{3}\pi r^3 + \pi r^2 h - 6) $.

The partial derivatives must be 0, so we have

$ 8\pi r + 2\pi h - \lambda (4 \pi r^2 + 2\pi r h) = 0 $
$ 2 \pi r - \lambda(\pi r^2) = 0 $

The second equation implies $ \lambda = \frac{2}{r} $.

Substitute this back to the first equation, solving get $ h = 0 $.

This seems reasonable because we know sphere maximize volume and minimize surface area, it only make sense if the cylindrical part vanish.

So it is simple now, $ \frac{4}{3}\pi r^3 = 6 $, $ r = \sqrt[3]{\frac{9}{2\pi}} = 1.127251652 $.

Farmer knew Lagrangian optimization!

Problem:



Solution:

Minimize x + 2y subject to xy = 405000

The Lagrangian is

$ L(x, y, \lambda) = x + 2y + \lambda(xy - 405000) $

The partial derivatives must be 0, so we get

$ 1 + \lambda y = 0 $
$ 2 + \lambda x = 0 $

Multiply the first equation by $ x $ and multiply the second equation by $ y $ gives $ x - 2y = 0 $.

In other words, $ x = 2y $, so $ xy = 2y^2 = 405000 $, $ y = 450, x = 900 $.

Limit

Problem:



Solution:

$ \frac{x^n - a^n}{x-a} = \frac{(x - a)(x^{n - 1} + ax^{n-2} + .. + a^{n-1})}{x - a} =  (x^{n - 1} + ax^{n-2} + .. + a^{n-1}) $

Therefore $ \lim\limits_{x \to a}\frac{x^n - a^n}{x-a} = \lim\limits_{x \to a} (x^{n - 1} + ax^{n-2} + .. + a^{n-1}) = na^{n-1} $.

Population

Problem:



Solution:

a) $ f(t) = 30 \times e^0.017t $
b) 41.43796053
c) 1.017145322
d) times

Finance 5

Problem:



Solution:

a)

Times	APY (Decimal)	APY (Percentage)
1	0.052	5.2%
12	0.053257411	5.325741057%
365	0.053371841	5.337184107%
Continuously	0.053375743	5.337574251%

b) 26912.4424

Finance 4

Problem:



Solution:

a) 2347.021742
b) $ f(t) = 2000 \times e^{\frac{3.2}{100}} $
c) $ e^{\frac{3.2}{100}} - 1 = 0.032517505 = 3.251750531\%$
d) $ (e^{\frac{3.2}{100}})^{10} - 1 = 0.377127764 = 37.71277643 \%$

Finance 3

Problem:



Solution:

a) 10326.85819
b) 2055.086475
c) 2937.22456

Finance 2

Problem:



Solution:

Year	Monthly	Daily	Continuously
1	4825.305364	4826.254424	4826.286816
2	5174.127079	5176.162615	5176.232095
4	5949.24245	5953.924315	5954.084156
10	9043.476195	9061.279018	9061.887184
30	36524.23864	36740.36641	36747.76461
APY	7.229008086%	7.250098317%	7.250818125%

To be continued with part b when the choices are available.