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

The tangent of f(x)=-2x^3,\at x=-1 is y=-6x-4

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

tangent of

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

y = - 6 x - 4

Solution steps

Find the tangent point:

(- 1, 2)

Find the slope of

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

Compute the slope of

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

x = - 1 : m = - 6

Find the line with slope m=

- 6

and passing through

(- 1, 2) : y = - 6 x - 4

y = - 6 x - 4

(1, - \frac{1}{2}), (- 2, - \frac{7}{2})

\int_{0}^{\frac{π}{2}} x^{3} cos (2 x) d x

\frac{d}{d x} (\frac{tan ( 3 x ^{3} + x ^{2} )}{e ^{x}})^{3}

x \to 2 lim (- x^{2} + 2 x - 6)

\int \frac{1}{a - x ^{2}} d x

What is the tangent of f(x)=-2x^3,\at x=-1 ?
The tangent of f(x)=-2x^3,\at x=-1 is y=-6x-4