What is the tangent of f(x)=2x^3-8x,(2,0) ?

The tangent of f(x)=2x^3-8x,(2,0) is y=16x-32

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tangent of f(x)=2x^3-8x,(2,0)

tangent of

f (x) = 2 x^{3} - 8 x, (2, 0)

y = 16 x - 32

Solution steps

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

(2, 0) : m = 16

Find the line with slope m=

16

and passing through

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

y = 16 x - 32

x - sin (x)

\frac{d}{d x} (ln (sin (10 x)))

y^{^{''}} + 6 y^{^{'}} - 16 y = - 272 sin (8 x)

\frac{d}{d x} (\frac{ln ( cos ( 2 x ) )}{x})

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

What is the tangent of f(x)=2x^3-8x,(2,0) ?
The tangent of f(x)=2x^3-8x,(2,0) is y=16x-32