Решите уравнение: 1. f'(x)·g'(x)=0 f(x)=x³-3x² g(x)=2/3√x 2. f(x)=(1+2x)(2x-1) ...

1) f'(x)*g'(x) = 0

$f'(x) = 3x^2-6x$

$g'(x) = -\frac{1}{3\sqrt{x^3}}$

$(3x^2-6x)(-\frac{1}{3}\frac{1}{\sqrt{x^3}})=0$

$\frac{6x-3x^2}{3x^{\frac{3}{2}}}=0$

$\frac{3x(2-x)}{3x\sqrt{x}} =0$

$x\neq0$

x = 2

2) $f(x) = (1+2x)(2x-1) = 4x^2-1$

$f'(x) = 8x$

$f'(-2) = -16$

3) $f(x) = 3 + \frac{x}{\sqrt{x}} = 3+\sqrt{x}$

$f'(x) = -\frac{1}{2\sqrt{x}}$

$f'(8) = -\frac{1}{2\sqrt{8}}=-\frac{1}{4\sqrt{2}}$

4) g(x) = 4sinx

g'(x) = 4cosx

$g'(-\frac{\pi}{3}) = 4cos(-\frac{\pi}{3}) = 4*\frac{1}{2} = 2$