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

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

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tangent of f(x)=x^3

tangent of

f (x) = x^{3}

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

f (x) = x^{3} : \frac{df ( x )}{d x} = 3 x^{2}

Compute the slope of

f (x) = x^{3}

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

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

2 sin (x)

y = 2 x - 4

θ = \frac{π}{3}

f (x) = x^{2} + x

x = - 3

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