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

The tangent of y=4x^3-2x,(-1,-2) is y=10x+8

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tangent of y=4x^3-2x,(-1,-2)

tangent of

y = 4 x^{3} - 2 x, (- 1, - 2)

y = 10 x + 8

Solution steps

Find the slope of

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

Compute the slope of

y = 4 x^{3} - 2 x

(- 1, - 2) : m = 10

Find the line with slope m=

10

and passing through

(- 1, - 2) : y = 10 x + 8

y = 10 x + 8

x^{2} y^{^{'}} = - 2 x y + (y)^{2}

f (x) = 3 x^{2}, at x = 10

\int_{1}^{10} \frac{37}{x ^{3}} d x

\frac{d}{d x} (ln (\frac{2 x + 1}{2 x - 1}))

x \to a lim (1 - 1 - x^{2})

What is the tangent of y=4x^3-2x,(-1,-2) ?
The tangent of y=4x^3-2x,(-1,-2) is y=10x+8