What is the tangent of x^3-3x+1 ?

The tangent of x^3-3x+1 is y=(3a_{0}^2-3)x-2a_{0}^3+1

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tangent of x^3-3x+1

tangent of

x^{3} - 3 x + 1

y = (3 a_{0}^{2} - 3) x - 2 a_{0}^{3} + 1

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} + 1)

Find the slope of

y = x^{3} - 3 x + 1 : \frac{d y}{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} + 1) : y = (3 a_{0}^{2} - 3) x - 2 a_{0}^{3} + 1

y = (3 a_{0}^{2} - 3) x - 2 a_{0}^{3} + 1

\int (sin (π x)) d x

\frac{\partial}{\partial y} (5 z e^{x yz})

\int e^{- x} sin (2 x) d x

\frac{\partial}{\partial x} (e^{x y} tan (y))

h \to 0 lim (\frac{( 18 h + 1 - 1 )}{h})

What is the tangent of x^3-3x+1 ?
The tangent of x^3-3x+1 is y=(3a_{0}^2-3)x-2a_{0}^3+1