What is the tangent of f(x)=2x^3-2x,\at x=1 ?

The tangent of f(x)=2x^3-2x,\at x=1 is y=4x-4

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tangent of f(x)=2x^3-2x,\at x=1

tangent of

f (x) = 2 x^{3} - 2 x, at x = 1

y = 4 x - 4

Solution steps

Find the tangent point:

(1, 0)

Find the slope of

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

Compute the slope of

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

x = 1 : m = 4

Find the line with slope m=

4

and passing through

(1, 0) : y = 4 x - 4

y = 4 x - 4

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

\frac{\partial}{\partial x} (\frac{8 x}{x ^{2} + y ^{2}})

\frac{\partial}{\partial b} (\frac{a ^{3} b ^{4}}{c ^{2} d ^{4}})

\frac{\partial}{\partial x} (- e^{- x - y})

\int_{\frac{1}{2}}^{1} (x^{- 3} - 24) d x

What is the tangent of f(x)=2x^3-2x,\at x=1 ?
The tangent of f(x)=2x^3-2x,\at x=1 is y=4x-4