Question

построить график функции y= Sin(arcSin 1/x)

Answer 1

1.

$-1leqfrac{1}{x}leq1, xneq0,\ left { {{frac{1}{x}geq-1} atop {frac{1}{x}leq1;}} right. left { {{xleq-1,} atop {xgeq1;}} right. \, D_y=(-infty;-1]cup[1;+infty);$

2.

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

3.

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

четная

4.

$y'=cos(arcsinfrac{1}{x})cdot(arcsinfrac{1}{x})'=cos(arcsinfrac{1}{x})cdotfrac{1}{sqrt{1-x^2}}cdot(frac{1}{x})'= \ =cos(pmarccossqrt{1-frac{1}{x^2}})cdotfrac{1}{sqrt{1-frac{1}{x^2}}}cdot(-frac{1}{x^2})= \ =-cos(arccossqrt{1-frac{1}{x^2}})cdotfrac{1}{x^2sqrt{1-frac{1}{x^2}}}=-sqrt{1-frac{1}{x^2}}cdotfrac{1}{x^2sqrt{1-frac{1}{x^2}}}= \ =-frac{1}{x^2}, \ 1-frac{1}{x^2}neq0, xneq-1, xneq1, \ -frac{1}{x^2}<0, y'<0, xin D_y, ysearrow;$

(-1;-1), (1;1) - точки разрыва.