What is the tangent of f(x)=x^3,\at x=6 ?

The tangent of f(x)=x^3,\at x=6 is y=108x-432

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tangent of f(x)=x^3,\at x=6

tangent of

f (x) = x^{3}, at x = 6

y = 108 x - 432

Solution steps

Find the tangent point:

(6, 216)

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 = 6 : m = 108

Find the line with slope m=

108

and passing through

(6, 216) : y = 108 x - 432

y = 108 x - 432

\int - 9 x d x

x \to 2 lim (51 - x - 7)

f (x) = x - 1, 1, 2

sin (5 t - 10)

\int_{2}^{5} x^{4} d x

What is the tangent of f(x)=x^3,\at x=6 ?
The tangent of f(x)=x^3,\at x=6 is y=108x-432