Question

Решите уравнение:1. f'(x)·g'(x)=0 f(x)=x³-3x² g(x)=2/3√x2. f(x)=(1+2x)(2x-1) f'(-2)3.φ(x)=3+x/√x φ'(8)4. g(x)=4sinx g'(-π/3)

Answer 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

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