Уравнение находится на фотографии

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

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

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

Уравнение находится ** фотографии