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

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

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tangent of x^3-3x+3

tangent of

x^{3} - 3 x + 3

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

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} + 3)

Find the slope of

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

Compute the slope of

y = x^{3} - 3 x + 3

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

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

\frac{8 x}{x ^{2} + 3}

n = 0 \sum \infty (\frac{1}{7})^{n}

\frac{\partial}{\partial x} (\frac{3 x}{5 y})

(6 x^{2} y^{2}) d x + (4 x^{3} y) d y = 0

\frac{d y}{d x} = \frac{9 x}{cos ( y )}

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