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

The tangent of f(x)= 1/(x^3),\at x=-2 is y=-3/16 x-1/2

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

tangent of

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

y = - \frac{3}{16} x - \frac{1}{2}

Solution steps

Find the tangent point:

(- 2, - \frac{1}{8})

Find the slope of

f (x) = \frac{1}{x ^{3}} : \frac{df ( x )}{d x} = - \frac{3}{x ^{4}}

Compute the slope of

f (x) = \frac{1}{x ^{3}}

x = - 2 : m = - \frac{3}{16}

Find the line with slope m=

- \frac{3}{16}

and passing through

(- 2, - \frac{1}{8}) : y = - \frac{3}{16} x - \frac{1}{2}

y = - \frac{3}{16} x - \frac{1}{2}

\int_{- 2}^{6} 4 y - (y^{2} - 12) d y

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

\frac{\partial}{\partial x} (40 x y^{- 4} z)

\int_{- 2}^{2} (4 - x^{2})^{2} d x

\frac{d}{d x} ((5 x - 3)^{4} (2 x - 7)^{- 3})

What is the tangent of f(x)= 1/(x^3),\at x=-2 ?
The tangent of f(x)= 1/(x^3),\at x=-2 is y=-3/16 x-1/2