What is the tangent of f(x)=x(27x^{-2}+2),\at x=-3 ?

The tangent of f(x)=x(27x^{-2}+2),\at x=-3 is y=-x-18

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tangent of f(x)=x(27x^{-2}+2),\at x=-3

tangent of

f (x) = x (27 x^{- 2} + 2), at x = - 3

y = - x - 18

Solution steps

Find the tangent point:

(- 3, - 15)

Find the slope of

f (x) = x (27 x^{- 2} + 2) : \frac{df ( x )}{d x} = - \frac{27}{x ^{2}} + 2

Compute the slope of

f (x) = x (27 x^{- 2} + 2)

x = - 3 : m = - 1

Find the line with slope m=

- 1

and passing through

(- 3, - 15) : y = - x - 18

y = - x - 18

\frac{\partial}{\partial x} (\frac{x - 1}{x + 1})

\int_{0}^{\infty} \frac{1}{e ^{x} - 1} d x

\frac{d}{d t} (\frac{1}{6} t^{3})

\int (x x^{2} + 3) d x

y^{^{''}} + 36 y = - 60 sec (6 t)

What is the tangent of f(x)=x(27x^{-2}+2),\at x=-3 ?
The tangent of f(x)=x(27x^{-2}+2),\at x=-3 is y=-x-18