What is the tangent of f(x)=x^3-4x ?

The tangent of f(x)=x^3-4x is y=(3a_{0}^2-4)x-2a_{0}^3

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tangent of f(x)=x^3-4x

tangent of

f (x) = x^{3} - 4 x

y = (3 a_{0}^{2} - 4) 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} - 4 a_{0})

Find the slope of

f (x) = x^{3} - 4 x : \frac{df ( x )}{d x} = 3 x^{2} - 4

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} - 4

Find the line with slope m=

3 a_{0}^{2} - 4

and passing through

(a_{0}, a_{0}^{3} - 4 a_{0}) : y = (3 a_{0}^{2} - 4) x - 2 a_{0}^{3}

y = (3 a_{0}^{2} - 4) x - 2 a_{0}^{3}

\frac{d}{d x} (\frac{sin ( x ^{2} )}{3 x})

\int x e^{x} d x

2 x y^{^{'}} + y = 6 x, y (4) = 26

f (x) = \frac{1}{x}

\int (x + 2) ln (x + 2) 2 d x

What is the tangent of f(x)=x^3-4x ?
The tangent of f(x)=x^3-4x is y=(3a_{0}^2-4)x-2a_{0}^3