Question

Хелп ми с алгеброй, pls. Дам 100 баллов

Answer 1

Объяснение:

№1.

в)

$f(x)=-x^2+2x\\\\f'(x)=(-x^2+2x)'=-2x+2.$

г)

$\displaystyle\\f(x)=\frac{x^2}{2} -9x\\\\f'(x)=(\frac{x^2}{2} -9x)'=\frac{2x}{2}-9=x-9.$

№2

$\displaystyle\\f'(x)=1\ \ \ \ f(x)=\frac{3x}{x-1} \\\\f'(x)=(\frac{3x}{x-1})'=\frac{(3x)'*(x-1)-3x*(x-1)'}{(x-1)^2}=\frac{3*(x-1)-3x*1}{(x-1)^2}= \\\\=\frac{3x-3-3x}{(x-1)^2}=\frac{-3}{(x-1)^2} .\\\\\frac{-3}{(x-1)^2} =1\\\\-3=1*(x-1)^2\\\\\ (x-1)^2=-3\\\\(x-1)^2\geq 0\ \ \ \ \Rightarrow\\\\x\in\varnothing.$

№3

$\displaystyle\\f'(x)\leq 0\ \ \ f(x)=\frac{2x^2+6}{3(x+1)} =\\\\ f'(x)=(\frac{2x^2+6}{3x+3})'=\frac{(2x^2+6)'*(3x+3)-(2x^2+6)*(3x+3)'}{(3x+3)^2}=\\\\=\frac{4x*(3x+3)-(2x^2+6)*3}{(3x+3)^2} =\frac{12x^2+12x-6x^2-18}{(3x+3)^2} =\frac{6x^2+12x-18}{(3*(x+1))^2} =\\\\=\frac{6*(x^2+2x-3)}{9*(x+1)^2} =\frac{2*(x^2+3x-x-3)}{3(x+1)^2}=\frac{2((x*(x+3)-(x+3))}{3*(x+1)^2}=\\\\ =\frac{2*(x+3)*(x-1)}{3*(x+1)^2}.$

$\displaystyle\\\frac{2*(x+3)*(x-1)}{3*(x+1)^2}\leq 0\ |*\frac{3}{2} \\\\\frac{(x+3)*(x-1)}{(x+1)^2} \leq 0$

-∞__+__-3__-__[-1]__-__1__+__+∞         ⇒

$x\in[-3;-1)U(-1;1].$