What is the tangent of y=4x^3-4x,(-1,0) ?

The tangent of y=4x^3-4x,(-1,0) is y=8x+8

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tangent of y=4x^3-4x,(-1,0)

tangent of

y = 4 x^{3} - 4 x, (- 1, 0)

y = 8 x + 8

Solution steps

Find the slope of

y = 4 x^{3} - 4 x : \frac{d y}{d x} = 12 x^{2} - 4

Compute the slope of

y = 4 x^{3} - 4 x

(- 1, 0) : m = 8

Find the line with slope m=

8

and passing through

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

y = 8 x + 8

x \to - 2 lim (- 3)

\int_{0}^{1} (\frac{1}{x ^{2}}) d x

\int cot^{3} (\frac{x}{2}) csc^{2} (\frac{x}{2}) d x

\frac{d}{d x} (\frac{( x ^{2} - 2 x ) ^{3}}{5 x})

\int 8 x^{3} e^{4 x} d x

What is the tangent of y=4x^3-4x,(-1,0) ?
The tangent of y=4x^3-4x,(-1,0) is y=8x+8