What is the tangent of f(x)= 9/(x^2),\at x=-1 ?

The tangent of f(x)= 9/(x^2),\at x=-1 is y=18x+27

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tangent of f(x)= 9/(x^2),\at x=-1

tangent of

f (x) = \frac{9}{x ^{2}}, at x = - 1

y = 18 x + 27

Solution steps

Find the tangent point:

(- 1, 9)

Find the slope of

f (x) = \frac{9}{x ^{2}} : \frac{df ( x )}{d x} = - \frac{18}{x ^{3}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 18

Find the line with slope m=

18

and passing through

(- 1, 9) : y = 18 x + 27

y = 18 x + 27

What is the tangent of f(x)= 9/(x^2),\at x=-1 ?
The tangent of f(x)= 9/(x^2),\at x=-1 is y=18x+27