What is the tangent of f(x)=2x^3-8x,\at x=2 ?

The tangent of f(x)=2x^3-8x,\at x=2 is y=16x-32

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tangent of f(x)=2x^3-8x,\at x=2

tangent of

f (x) = 2 x^{3} - 8 x, at x = 2

y = 16 x - 32

Solution steps

Find the tangent point:

(2, 0)

Find the slope of

f (x) = 2 x^{3} - 8 x : \frac{df ( x )}{d x} = 6 x^{2} - 8

Compute the slope of

f (x) = 2 x^{3} - 8 x

x = 2 : m = 16

Find the line with slope m=

16

and passing through

(2, 0) : y = 16 x - 32

y = 16 x - 32

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

y = x^{4} + 2 x^{2} - x

\frac{d}{d x} (\frac{1}{5 x ^{4}} - \frac{5}{2 x})

f (x) = 6 x^{2}, (2, 24)

\int 1 - 4 y d y

What is the tangent of f(x)=2x^3-8x,\at x=2 ?
The tangent of f(x)=2x^3-8x,\at x=2 is y=16x-32