Question

Решите уравнение:

 

1. f'(x)·g'(x)=0

    f(x)=x³-3x²

    g(x)=2/3√x

 

2. 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

$f'(x) = 3x^2-6x$ 

$g'(x) = -frac{1}{3sqrt{x^3}}$ 

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

$frac{6x-3x^2}{3x^{frac{3}{2}}}=0$ 

$frac{3x(2-x)}{3xsqrt{x}} =0$ 

$xneq0$ 

x = 2 

2) $f(x) = (1+2x)(2x-1) = 4x^2-1$

$f'(x) = 8x$ 

$f'(-2) = -16$ 

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

$f'(x) = -frac{1}{2sqrt{x}}$ 

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

4) g(x) = 4sinx

g'(x) = 4cosx

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