What is the tangent of x^3-27x ?

The tangent of x^3-27x is y=(3a_{0}^2-27)x-2a_{0}^3

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tangent of x^3-27x

tangent of

x^{3} - 27 x

y = (3 a_{0}^{2} - 27) 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} - 27 a_{0})

Find the slope of

y = x^{3} - 27 x : \frac{d y}{d x} = 3 x^{2} - 27

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} - 27

Find the line with slope m=

3 a_{0}^{2} - 27

and passing through

(a_{0}, a_{0}^{3} - 27 a_{0}) : y = (3 a_{0}^{2} - 27) x - 2 a_{0}^{3}

y = (3 a_{0}^{2} - 27) x - 2 a_{0}^{3}

x^{2} y^{^{'}} - y x = 0

\frac{d}{d x} (6 x^{4} - 2 x + 7)

\frac{1}{( s - 1 ) ^{2}}

\int_{0}^{1} k x d x

y = cos (x^{2})

What is the tangent of x^3-27x ?
The tangent of x^3-27x is y=(3a_{0}^2-27)x-2a_{0}^3