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

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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

Compute the slope of

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

x = a_{0} : 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})

x \to o + lim ((x + 1)^{in x})

x \to \infty lim (x e^{- s x})

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

\int \frac{x ^{2} + 10 x + 2}{x ^{3} + 15 x ^{2} + 6 x + 9} d x

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