MaxWong

You don't. That's not how you do it.

You finally get one side as a polynomial of x, with known coefficients, another side as a polynomial of x, with unknown coefficients.

You compare the coefficient of each term, then you can find the unknowns.

$\dfrac{x+7}{x^2-x-2} = \dfrac{A}{x-2}+\dfrac B{x+1}\\ \dfrac{x+7}{x^2-x-2} = \dfrac{A(x+1) + B(x-2)}{x^2-x-2}\\ \text{As the denominators are the same, so the numerators must be the same.}\\ x + 7 = A(x+1) + B(x-2)\\ x + 7 = Ax + A + Bx - 2B\\ x + 7 = (A+B)x + (A-2B)\\ \begin{cases} A + B = 1---(1)\\ A - 2B = 7---(2)\\ \end{cases}\\ (1)-(2):\\3B = -6\\ B = -2\\ A - 2 = 1 \implies A = 3\\ \therefore (A,B) = (3,-2)$

Factored/Simplified Expressions

You take an expression, find all its factors, write it out as a product of factors, you get the factored form.

You expand all the product, then you get the simplified form.

Example:

$x^2+2x-15\qquad \text{This is}\color{blue}{\text{ simplified}}\color{black}\text{.}\\ (x+5)(x-3)\qquad \text{This is}\color{magenta}{\text{ factored}}\color{black}\text{.}\\$

Removable discontinuity

This usually only happens when the given function is rational.

1. Find the G.C.D. of the numerator and the denominator.

2. Set the G.C.D. to 0

3. The roots of the equation in step 2 is where the removable discontinuity occurs.

Example:

Find the removable discontinuity(discontinuities) of the function $f(x) = \dfrac{\dfrac{1}{4}x^2-x+1}{x^2-4}$.

Solution:

$f(x)\\ =\dfrac{\dfrac{1}{4}x^2-x+1}{x^2-4}\\ =\dfrac{\dfrac{1}{4}(x^2-4x+4)}{(x-2)(x+2)}\\ =\dfrac{\dfrac{1}{4}(x-2)^2}{(x-2)(x+2)}\\ =\dfrac{(x-2)^2}{4(x-2)(x+2)}$

GCD of numerator and denominator is (x-2). <- Set this to 0.

x - 2 = 0

x = 2

The removable discontinuity of this function occurs at x = 2.

Vertical asymptotes and horizontal asymptotes

Vertical asymptotes occurs at (unremovable) discontinuities.

Horizontal asymptotes is just a horizontal line which the function approaches when x reaches positive or negative infinity.

How to find...

Vertical asymptotes? Find all the discontinuities, exclude the removable ones and the "broken" points(like this one below, at x = 1.)

Horizontal asymptotes? That's a calculus thing. If you want to do it the algebra way, you can substitute larger and larger numbers into the function and see if the function reaches a value eventually, and then repeat with negative numbers.

Example:

Find all the horizontal and vertical asymptotes of the function y = 1/x.

Solution:

Let's find the horizontal asymptotes first. If you substitute large numbers, you will find that it approaches 0 eventually. Same for negative numbers. So the only horizontal asymptote, in this case, is y = 0.

The function has a discontinuity at x = 0, and it is not removable. So vertical asymptote is at x = 0.

Domain and range

Domain is the set of possible inputs of the function.

Range is the set of possible outputs of the function.

Example:

Find the domain and range of the function $f(x) = \dfrac{1}{\sqrt{x-5}}$.

Solution:

This requires some kind of sense of mathematics. Firstly, the denominator can't be 0, or it will give infinity. So we have $\sqrt{x-5} \neq 0 \implies x - 5 \neq 0$.

Also, $\sqrt{ }$ can't take negative numbers as input. So $x - 5 \geq 0$. But actually as the denominator can't be 0, x - 5 can't be 0. So $x - 5 > 0\implies x > 5$.

So the domain of the function $\text{D}\{ f(x)\}$ is $(5,\infty)$.

Now let's consider the output of the function. $\sqrt{ }$ only gives out nonnegative numbers(0 and positive numbers if you don't know what are nonnegative numbers), So $\sqrt{x-5} \ge 0 \implies f(x) = \dfrac{1}{\sqrt{x-5}} \le \infty$, which doesn't make any sense. But $f(x) = \dfrac{1}{\text{a nonnegative number}}$, considering that, we get the range of f(x) as the set of all positive real numbers.

$\text{R}\{f(x)\}$ is $(0,\infty)$.

P.S.: How do I know these without knowing all the terminologies? Google it ( ͡° ͜ʖ ͡°).

$\cos x \sin x + \sin x = 0\\ \sin x \left(\cos x + 1\right) = 0\\ \sin x = 0 \vee \cos x = -1\\ x = 0^{\circ} \vee x = 180^{\circ} \vee x = 360^{\circ}$

The big V means "or"

Actually, (n)(2pi), not (2n)(pi). After n periods, which is n(2pi), the value of the function is still the same.

You can derive this from a right triangle.

$\displaystyle{ \sum_{r=k+1}^{60} (3r+2)\log_8\left(4^r\right)\\ =\sum_{r=k+1}^{60} \left(2r^2 + \dfrac{4}{3}r\right)\\ =2\sum_{r=k+1}^{60}r^2 + \dfrac{4}{3}\sum_{r=k+1}^{60} r\\ =2\left(73810 - \dfrac{k(k+1)(2k+1)}{6}\right)+\dfrac{4}{3}\left(1830-\dfrac{k(k+1)}{2}\right)\\ =150060-\dfrac{1}{3}\left(2k^3+3k^2+k\right)-\dfrac{2}{3}\left(k^2+k\right)\\ =150060-\dfrac{2}{3}k^3-\dfrac{5}{3}k^2-k\\ 150060-\dfrac{2}{3}k^3-\dfrac{5}{3}k^2-k > 106060\\ 2k^3+5k^2+3k-132000<0\\ k < 39.58 }$

Greatest value of k is 39.

$\log_8\left(4^r\right)\\ =r \log_8\left(4\right)\\ =\dfrac{2}{3} r$

c.

GD = $\dfrac{\text{BD}}{2} \approx 3.24\text m$ 

Entire height of the house is $\text{GD} \tan\left(50^{\circ}\right) + 6\text m= 9.86\text m$

b.

Construct line DP such that DP is perpendicular to HE and P is on line HE.

PE = $\dfrac{3\text m}{\tan\left(50^{\circ}\right)} \approx 2.52\text m$

BD = HE - PE = (9 - 2.52)m = 6.48m which is around 6.5 m.

a.

BH = 6m - 3m = 3m

DE = $\dfrac{3\text m}{\sin\left(50^{\circ}\right)} \approx 3.92\text m$