MaxWong

\(\sqrt{t} \ge 0 \quad\forall t \in D(\sqrt x),\text{ where }D(\sqrt x) \text{ denotes the domain of }f(x) = \sqrt x.\)

Whenever you see a square root, you can be sure that it is greater than or equal to 0.

Now write f(x) > 0 using interval notation.

(a) Think about what makes the fraction undefined. What value(s) of x makes the denominator 0?

(b) Notice that the function has a vertical asymptote. Substitute points around the vertical asymptote to find out if the range is \((-\infty,\infty)\), \((-\infty ,0)\),\((0,\infty)\),\([0,\infty)\), or \((-\infty ,0]\).

(c) Notice that the function has a vertical asymptote. Substitute points around the vertical asymptote to find out if the range is \((-\infty,\infty)\), \((-\infty ,0)\),\((0,\infty)\),\([0,\infty)\), or \((-\infty ,0]\)​.

418 or 480? 

And did you mean that whether 3 integers (a, b, and c) can be found such that \(a< b< c<180\) and \(a + b + c = 418\text{(or 480)}\)?

Nice. Is this a linear search on the set of integers?

\(\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{12}\\ \dfrac{x+y}{xy} = \dfrac{1}{12}\\ \text{Let }x + y = k, xy = 12k.\\ y(k-y) = 12k\\ -y^2 +ky = 12k\\ y^2 -ky+12k = 0\\ y = \dfrac{k+\sqrt{k^2 -48k}}{2}\\ \implies k^2 - 48k \text{ is a perfect square.}\\ k^2 - 48k \ge 0\\ k^2 - 48k = t^2\\ k \le 0\text{(rejected)} \vee k\ge48\\ k \ge 48\\ k \text{ cannot be 48. If } k = 48\text{, } x = y = 24.\\ k \ge 49\\ k = 49 \text{ satisfies the condition that } k^2 - 48k \text{ is a perfect square.}\\ \min(k) = 49\\ \boxed{\min(x + y) = \color{red}49}\)

4)

\(f(x) = \log \left(x + \sqrt{1 +x^2}\right)\\ \begin{array}{rcll} f(-x) &=& \log \left(-x + \sqrt{1 + x^2}\right)\\ &=&\log \left(\dfrac{(-x+\sqrt{1+x^2})(x + \sqrt{1+x^2})}{x+\sqrt{1+x^2}}\right)\\ &=&\log \left(\dfrac{(\sqrt{1+x^2})^2-x^2}{x+\sqrt{1+x^2}}\right)\\ &=&\log \left(\dfrac{1}{x+\sqrt{1+x^2}}\right)\\ &=&\log \left(\left(x + \sqrt{1+x^2}\right)^{-1}\right)\\ &=&-\log\left(x+\sqrt{1+x^2}\right)\\ &=&-f(x)\\ \end{array}\\ \therefore \text{It is an odd function.}\)

This sum is obviously telescoping :)

\(\begin{array} {rll} \phantom{=}&\displaystyle\sum^{9999}_{n=1}\dfrac{1}{\left(\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt[4]{n}+\sqrt[4]{n+1}\right)}\\ =&\displaystyle \sum^{9999}_{n=1}\dfrac{\color{red}{\sqrt[4]{n+1}-\sqrt[4]{n}}}{\left(\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt[4]{n}+\sqrt[4]{n+1}\right)\color{red}{\left(\sqrt[4]{n+1}-\sqrt[4]{n}\right)}}\\ =&\displaystyle \sum^{9999}_{n=1}\dfrac{\sqrt[4]{n+1}-\sqrt[4]{n}}{\left(\sqrt{n}+\sqrt{n+1}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}\\ =&\displaystyle \sum^{9999}_{n=1}\dfrac{\sqrt[4]{n+1}-\sqrt[4]{n}}{n+1 - n}\\ =&\displaystyle \sum^{9999}_{n=1}\left(\sqrt[4]{n+1}-\sqrt[4]{n}\right)\\ =&\displaystyle (\color{blue}{\sqrt[4]{2}}\color{black} - \sqrt[4]1) + (\color{green}{\sqrt[4]{3}}\color{black} - \color{blue}\sqrt[4]2\color{black}) + (\sqrt[4]4 - \color{green}\sqrt[4]3\color{black}) + \cdots+(\sqrt[4]{10000}-\sqrt[4]{9999})\\ =&\displaystyle \sqrt[4]{10000} - \sqrt[4]1\\ =&\displaystyle 10 - 1\\ =& 9 \end{array}\)

\(\begin{cases} 2x + 5y = -8\\ 6x + 15y = 16 + a \end{cases}\\ \begin{cases} 2x + 5y = -8\\ 2x + 5y = \dfrac{16 + a}3 \end{cases}\\ \dfrac{16+a}{3} = -8\\ 16 + a = -24\\ a = -40\)

\(P(x) = x^3 + bx^2 + cx + d \\ (2^{1/3}+1)^3 + b(2^{1/3}+1)^2 + c(2^{1/3}+1) +d = 0\\ 3 + 3\left(2^{2/3}+2^{1/3}\right) + b(2^{2/3})+2b(2^{1/3})+b+c(2^{1/3})+c+d=0\\ (3+b)(2^{2/3}) + (3+2b+c)(2^{1/3}) + (3+b+c+d) = 0\\ 3 + b = 0\\ b = -3\\ 3 + 2b + c = 0\\ 3 - 6 + c = 0\\ c = 3\\ 3 + b + c + d = 0\\ 3 - 3 + 3 + d = 0\\ d = -3\\ \therefore \text{The required monic polynomial is } x^3 -3x^2+3x-3=0 \)