What is the tangent of x^3+2 ?

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

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tangent of x^3+2

tangent of

x^{3} + 2

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{3} + 2)

Find the slope of

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

Compute the slope of

y = x^{3} + 2

x = a_{0} : m = 3 a_{0}^{2}

Find the line with slope m=

3 a_{0}^{2}

and passing through

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

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

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