What is the tangent of f(x)=x^3-3x+1 ?

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

tangent of f(x)=x^3-3x+1

tangent of

f (x) = 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

f (x) = x^{3} - 3 x + 1 : \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} + 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

n \to \infty lim (x^{2 n})

\frac{d}{d x} (\frac{x ^{2} - 16}{x - 4})

x \to 0 lim (tan (2 x) csc (4 x))

x \to π lim (\frac{5 x}{3})

\int y^{2} e^{- y} d y

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