What is the tangent of f(x)= 9/(x^2),\at x=3 ?

The tangent of f(x)= 9/(x^2),\at x=3 is y=-2/3 x+3

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tangent of f(x)= 9/(x^2),\at x=3

tangent of

f (x) = \frac{9}{x ^{2}}, at x = 3

y = - \frac{2}{3} x + 3

Solution steps

Find the tangent point:

(3, 1)

Find the slope of

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

Compute the slope of

f (x) = \frac{9}{x ^{2}}

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

Find the line with slope m=

- \frac{2}{3}

and passing through

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

y = - \frac{2}{3} x + 3

\frac{\partial}{\partial x} (e^{- 7 y} sin (7 x))

\frac{d}{d x} (\frac{4 x}{5 2} + 3)

25 t^{2}

f (t) = 7 - \frac{1}{3} t

\int \frac{7 x ^{2} + 5 x + 7}{( x ^{2} + 1 ) ^{2}} d x

What is the tangent of f(x)= 9/(x^2),\at x=3 ?
The tangent of f(x)= 9/(x^2),\at x=3 is y=-2/3 x+3