Question

Людии помогите пожалуйста очень надо
Дам 50 баллов

Answer 1

$175)\ \ g(x)=2cos^2x-4cosx+3\\\\g'(x)=2\cdot 2\, cosx\cdot (-sinx)+4\, sinx=-4\, cosx\cdot sinx+4sinx=-4sinx(cosx-1)\\\\-4\, sinx(cosx-1)=0\\\\a)\ \ sinx=0\ \ ,\ \ x=\pi n\ ,\ n\in Z\\\\b)\ \ cosx=1\ \ ,\ \ x=2\pi k\ ,\ k\in Z\ \ \Big(\ esli\ n=2k\ ,\ to\ \ x=\pi n=2\pi k\ ,\ k\in Z\ \Big)\\\\Otvet:\ \ x=\pi n\ ,\ n\in Z\ .$

$177)\ \ f(x)=sin^4(2x-7)\ \ \ ,\ \ \ \ (u^4)'=4u^3\cdot u'\\\\f'(x)=4\, sin^3(2x-7)\cdot (sin(2x-7))'=4\, sin^3(2x-7)\cdot cos(2x-7)\cdot2=\\\\=8\, sin^3(2x-7)\cdot cos(2x-7)\\\\\\b)\ \ f(x)=\sqrt{cos8x}\ \ ,\ \ \ (\sqrt{u})'=\dfrac{1}{2\, \sqrt{u}}\cdot u'\\\\f'(x)=\dfrac{1}{2\, \sqrt{cos8x}}\cdot (cos8x)'=\dfrac{1}{2\, \sqrt{cos8x}}\cdot (-sin8x\cdot 8)=-\dfrac{4\, sin8x}{\sqrt{cos8x}}$

$203)\ \ \ f(x)=\dfrac{x-2}{x^2+4}\\\\OX:\ \ y=0\ \ \to \ \ \dfrac{x-2}{x^2+4}=0\ \ ,\ \ \ x_0=2\\\\f(2)=0\ \ ,\\\\f'(x)=\dfrac{x^2+4-2x(x-2)}{(x^2+4)^2}=\dfrac{-x^2+4x+4}{(x^2+4)^2}\ \ ,\\\\\\f'(2)=\dfrac{-4+8+4}{(4+4)^2}=\dfrac{8}{8^2}=\dfrac{1}{8}\\\\\\y=0+\dfrac{1}{8}\, (x-2)\\\\\\\boxed {\ y=\dfrac{1}{8}\, x-\dfrac{1}{4}\ }$