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

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

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tangent of y=x^3-3x+1

tangent of

y = 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

Compute the slope of

y = 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

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

x \to - 2 + lim (2 - x)

\int x^{2} x + 5 d x

\int in y d y

(1 + x^{4}) \frac{d y}{d x} = \frac{x ^{3}}{y}

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