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2024-08-21

Full / hollow dot at the end of graphs

2024-08-23

$\sqrt{x^{2}} = | x |$

One to one: (of a function) has an inverse. Can be tested with the horizontal line test

\begin{aligned} f \circ f^{- 1} (x) & = x for dom f^{- 1} \\ f^{- 1} \circ f (x) & = x for dom f \end{aligned}

2024-08-27

\begin{aligned} 6 (3 - x)^{4} + 2 x (3 - x)^{3} - x^{2} (3 - x)^{2} & = A B^{2} C \\ (3 - x)^{2} (6 (3 - x)^{2} + 2 x (3 - x) - x^{2}) \\ (3 - x)^{2} (3 x^{2} - 30 x + 54) \\ 3 (3 - x)^{2} (x^{2} - 10 x + 18) \\ 3 (3 - x)^{2} ((x - 5)^{2} - 7) \\ 3 (3 - x)^{2} (x - 5 + \sqrt{7}) (x - 5 - \sqrt{7}) \end{aligned}

\begin{aligned} \frac{\frac{1}{\sqrt{17 (x + h)}} - \frac{1}{\sqrt{17 x}}}{h} \\ = & \frac{\sqrt{17 x} - \sqrt{17 (x + h)}}{17 h \sqrt{x (x + h)}} \\ = & \frac{(\sqrt{17 x} - \sqrt{17 (x + h)}) (\sqrt{17 x} + \sqrt{17 (x + h)})}{17 h \sqrt{x (x + h)} (\sqrt{17 x} + \sqrt{17 (x + h)})} \\ = & \frac{17 x - 17 (x + h)}{17 h (\sqrt{17 (x + h)} x + \sqrt{17 x} (x + h))} \\ = & \frac{1}{\sqrt{17} (\sqrt{x + h} x + \sqrt{x} (x + h))} \end{aligned}

\begin{aligned} 0 & = x^{2} - 20 x + 75 \\ 0 & = (x - 5) (x - 15) \\ x & = 5 or 15 \end{aligned}

\begin{aligned} \frac{(x^{2} - 3)^{2}}{4} & = 2 - x^{2} \\ x^{4} - 6 x^{2} + 9 & = 8 - 4 x^{2} \\ x^{4} - 2 x^{2} + 1 & = 0 \\ (x^{2} - 1)^{2} & = 0 \\ x = \pm 1 \end{aligned}

\begin{aligned} - \frac{1}{x - 3} + \frac{x + 5}{2 x + 1} & = 0 \\ - 2 - \frac{7}{x - 3} + x + 5 & = 0 \\ - 7 + (x + 3) (x - 3) & = 0 \\ x^{2} - 16 & = 0 \\ (x + 4) (x - 4) & = 0 \\ x & = \pm 4 \end{aligned}

\begin{aligned} (2 x + x^{2} (15 - 17 x)) (15 - 17 x)^{- \frac{1}{2}} & = 0 \\ - 17 x^{3} + 15 x^{2} + 2 x & = 0 \\ - 17 x^{2} + 15 x + 2 & = 0 \\ (- x + 1) (17 x + 2) & = 0 \\ x & = 1 or - \frac{2}{17} \end{aligned}

\begin{aligned} \frac{\frac{x}{4 x + 7} - \frac{a}{4 a + 7}}{x - a} \\ = & \frac{x (4 a + 7) - a (4 x + 7)}{(x - a) (4 x + 7) (4 a + 7)} \\ = & \frac{7 (x - a)}{(x - a) (4 x + 7) (4 a + 7)} \\ = & \frac{7}{(4 x + 7) (4 a + 7)} \end{aligned}

2024-08-28

Don't confuse $\sin^{- 1} x$ with $(\sin x)^{- 1}$
Prefer $\arcsin$ and $a s i n$ to $\sin^{- 1}$

2024-09-06

\begin{aligned} a = - 3 \\ (- 2)^{2} + b \cdot - 2 + c & = - 3 \\ - 2 b + c & = - 7 \\ 3^{2} + b \cdot 3 + c & = 22 \\ 3 b + c & = 13 \\ 5 b & = 20 \\ b & = 4 \\ - 2 \cdot 4 + c & = - 7 \\ c & = 1 \end{aligned}

2024-09-10

Factorization of cubes:

\begin{aligned} a^{3} - b^{3} & = (a - b) (a^{2} + a b + b^{2}) \\ a^{3} + b^{3} & = (a + b) (a^{2} - a b + b^{2}) \end{aligned}

\begin{aligned} lim_{x \to 1} \frac{\sqrt{2 + x^{2}} - \sqrt{4 x - x^{2}}}{x^{2} - 1} & = lim_{x \to 1} \frac{2 + x^{2} - (4 x - x^{2})}{(x^{2} - 1) (\sqrt{2 + x^{2}} + \sqrt{4 x - x^{2}})} \\ = lim_{x \to 1} \frac{2 x^{2} - 4 x + 2}{(x^{2} - 1) (\sqrt{2 + x^{2}} + \sqrt{4 x - x^{2}})} \\ = 2 lim_{x \to 1} \frac{(x - 1)^{2}}{(x - 1) (x + 1) (\sqrt{x^{2} + 2} + \sqrt{4 x - x^{2}})} \\ = 2 lim_{x \to 1} \frac{x - 1}{(x + 1) (\sqrt{x^{2} + 2} + \sqrt{4 x - x^{2}})} \\ = 0 \end{aligned}

2024-09-11

$lim_{x \to 1^{+}} \frac{\sqrt{x^{2} + x}}{\sqrt{x - x^{2}}}$

When $x = 1$ , limit = $\frac{\sqrt{2}}{0}$ .
We factorize $x - x^{2}$ to $x (1 - x)$ .
Now it becomes clear that $x (1 - x) = 0^{-}$ .
The limit diverges to $- \infty$ .

2024-10-01

Point-slope form: $y - y_{o} = m (x - x_{o})$

2024-10-09

Find $k^{'} (1)$ of $k (x) = (x^{2} + 3)^{3} (x - 2 x^{2})^{2}$ .

\begin{aligned} k^{'} (1) & = {\frac{d}{d x} k (x) |}_{x = 1} \\ = {\frac{d}{d x} ((x^{2} + 3)^{3} (x - 2 x^{2})^{2}) |}_{x = 1} \\ = {3 (x^{2} + 3)^{2} (x^{2} + 3)^{'} (x - 2 x^{2})^{2} + (x^{2} + 3)^{3} 2 (x - 2 x^{2}) (x - 2 x^{2})^{'} |}_{x = 1} \\ = {3 (x^{2} + 3)^{2} 2 x (x - 2 x^{2})^{2} + (x^{2} + 3)^{3} 2 (x - 2 x^{2}) (1 - 4 x) |}_{x = 1} \\ = 3 (4)^{2} \cdot 2 (1 - 2)^{2} + (4)^{3} \cdot 2 (1 - 2) (1 - 4) \\ = 3 \cdot 16 \cdot 2 \cdot 1 + 64 \cdot 2 \cdot - 1 \cdot - 3 \\ = 96 + 384 \\ = 480 \end{aligned}

Find the derivative of $h (y) = \tan (\sec ((y^{3} + 1)^{4}))$ .

\begin{aligned} h^{'} (y) & = \frac{d}{d y} h (y) \\ = \frac{d}{d y} (\tan (\sec ((y^{3} + 1)^{4}))) \\ = \sec^{2} (\sec ((y^{3} + 1)^{4})) (\sec ((y^{3} + 1)^{4}))^{'} \\ = \sec^{2} (\sec ((y^{3} + 1)^{4})) \sec ((y^{3} + 1)^{4}) \tan ((y^{3} + 1)^{4}) 4 (y^{3} + 1)^{3} 3 y^{2} \end{aligned}

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Ajitani Hifumi

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