Question

Уравнение находится на фотографии

Answer 1

1.

$f(x)' = x^3-12x\f(x)' = frac{d}{dx}left(x^3-12xright) = frac{d}{dx}left(x^3right)-frac{d}{dx}left(12xright)=3x^2-12 =\= 3(x^2-4) = 3(x-2)(x+2)\\3(x-2)(x+2) = 0\x=2,:: x=-2$

2.

$f(x)'=x^2-3x+1\f(x)'=frac{d}{dx}left(x^2-3x+1right) =frac{d}{dx}left(x^2right)-frac{d}{dx}left(3xright)+frac{d}{dx}left(1right) =2x-3\\2x-3=0\2x=3\x=1,5$

3.

$f(x)'=x^3-x^2-4x-3\f(x)'= frac{d}{dx}left(x^3-x^2-4x-3right)=\=frac{d}{dx}left(x^3right)-frac{d}{dx}left(x^2right)-frac{d}{dx}left(4xright)-frac{d}{dx}left(3right)=\= 3x^2-2x-4\\3x^2-2x-4=0\D = 4+48 = 52\x_{1,2} = frac{2pm sqrt{D} }{2cdot 3} \x_{1} = frac{2 + 2sqrt{13} }{2cdot 3} =frac{2(1+sqrt{13})}{2cdot 3} =frac{1+sqrt{13}}{3} approx 1,54\x_{2} = frac{2 - 2sqrt{13} }{2cdot 3} =frac{2(1-sqrt{13})}{2cdot 3} =frac{1-sqrt{13}}{3} approx -0,87$