What is the tangent of f(x)=4x^3-4x,\at x=1 ?

The tangent of f(x)=4x^3-4x,\at x=1 is y=8x-8

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tangent of f(x)=4x^3-4x,\at x=1

tangent of

f (x) = 4 x^{3} - 4 x, at x = 1

y = 8 x - 8

Solution steps

Find the tangent point:

(1, 0)

Find the slope of

f (x) = 4 x^{3} - 4 x : \frac{df ( x )}{d x} = 12 x^{2} - 4

Compute the slope of

f (x) = 4 x^{3} - 4 x

x = 1 : m = 8

Find the line with slope m=

8

and passing through

(1, 0) : y = 8 x - 8

y = 8 x - 8

\frac{d}{d t} (t^{2} + 5 t)

\frac{\partial}{\partial t} (ln (k x - m t))

f (x) = \frac{x ^{2} - 1}{x ^{2} + 1}

\frac{d}{d x} (\frac{4 x}{9 y})

\frac{d}{d x} (- 7 (3 x^{4} + 2)^{- 4})

What is the tangent of f(x)=4x^3-4x,\at x=1 ?
The tangent of f(x)=4x^3-4x,\at x=1 is y=8x-8