Question

Построить график функции y= sin(arcsin 1/x)

Answer 1

1.

$-1\leq\frac{1}{x}\leq1, x\neq0,\\ \left \{ {{\frac{1}{x}\geq-1} \atop {\frac{1}{x}\leq1;}} \right. \ \left \{ {{x\leq-1,} \atop {x\geq1;}} \right. \\, D_y=(-\infty;-1]\cup[1;+\infty);$

2.

$E_y=[-1;0)\cup(0;1]$

3.

$\sin(\arcsin\frac{1}{-x})=\sin(\arcsin(-\frac{1}{x}))=\sin(-\arcsin\frac{1}{x})= \\ =-\sin(\arcsin\frac{1}{x}), \\ f(-x)=-f(x);$

четная

4.

$y'=\cos(\arcsin\frac{1}{x})\cdot(\arcsin\frac{1}{x})'=\cos(\arcsin\frac{1}{x})\cdot\frac{1}{\sqrt{1-x^2}}\cdot(\frac{1}{x})'= \\ =\cos(\pm\arccos\sqrt{1-\frac{1}{x^2}})\cdot\frac{1}{\sqrt{1-\frac{1}{x^2}}}\cdot(-\frac{1}{x^2})= \\ =-\cos(\arccos\sqrt{1-\frac{1}{x^2}})\cdot\frac{1}{x^2\sqrt{1-\frac{1}{x^2}}}=-\sqrt{1-\frac{1}{x^2}}\cdot\frac{1}{x^2\sqrt{1-\frac{1}{x^2}}}= \\ =-\frac{1}{x^2}, \\ 1-\frac{1}{x^2}\neq0, x\neq-1, x\neq1, \\ -\frac{1}{x^2}$

(-1;-1), (1;1) - точки разрыва.