What is the tangent of f(x)=x^3-3x ?

The tangent of f(x)=x^3-3x is y=(3a_{0}^2-3)x-2a_{0}^3

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tangent of f(x)=x^3-3x

tangent of

f (x) = x^{3} - 3 x

y = (3 a_{0}^{2} - 3) x - 2 a_{0}^{3}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} - 3 a_{0})

Find the slope of

f (x) = x^{3} - 3 x : \frac{df ( x )}{d x} = 3 x^{2} - 3

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 3 a_{0}^{2} - 3

Find the line with slope m=

3 a_{0}^{2} - 3

and passing through

(a_{0}, a_{0}^{3} - 3 a_{0}) : y = (3 a_{0}^{2} - 3) x - 2 a_{0}^{3}

y = (3 a_{0}^{2} - 3) x - 2 a_{0}^{3}

\int 3 t^{2} sin (t) d t

\frac{\partial}{\partial x} (\frac{1}{\frac{1}{x} + \frac{1}{o}})

\frac{\partial}{\partial y} (3 x^{\frac{y}{z}})

\frac{\partial}{\partial y} (2 e^{- 2 x y} + x y^{3})

x, 20 - x^{2}

What is the tangent of f(x)=x^3-3x ?
The tangent of f(x)=x^3-3x is y=(3a_{0}^2-3)x-2a_{0}^3