What is the tangent of f(x)=x^3-6x,\at x=2 ?

The tangent of f(x)=x^3-6x,\at x=2 is y=6x-16

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tangent of f(x)=x^3-6x,\at x=2

tangent of

f (x) = x^{3} - 6 x, at x = 2

y = 6 x - 16

Solution steps

Find the tangent point:

(2, - 4)

Find the slope of

f (x) = x^{3} - 6 x : \frac{df ( x )}{d x} = 3 x^{2} - 6

Compute the slope of

f (x) = x^{3} - 6 x

x = 2 : m = 6

Find the line with slope m=

6

and passing through

(2, - 4) : y = 6 x - 16

y = 6 x - 16

f (x) = cos^{3} (2 + x)

\int e^{- x} sin (3 x) d x

h \to 0 lim ((x + h) 3 - x^{3} h)

\int (1 + y^{3})^{2} d y

y = \frac{3 + x f ( x )}{x}

What is the tangent of f(x)=x^3-6x,\at x=2 ?
The tangent of f(x)=x^3-6x,\at x=2 is y=6x-16