What is the tangent of f(x)= 1/(x^2),(3, 1/9) ?

The tangent of f(x)= 1/(x^2),(3, 1/9) is y=-2/27 x+1/3

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tangent of f(x)= 1/(x^2),(3, 1/9)

tangent of

f (x) = \frac{1}{x ^{2}}, (3, \frac{1}{9})

y = - \frac{2}{27} x + \frac{1}{3}

Solution steps

Find the slope of

f (x) = \frac{1}{x ^{2}} : \frac{df ( x )}{d x} = - \frac{2}{x ^{3}}

Compute the slope of

f (x) = \frac{1}{x ^{2}}

(3, \frac{1}{9}) : m = - \frac{2}{27}

Find the line with slope m=

- \frac{2}{27}

and passing through

(3, \frac{1}{9}) : y = - \frac{2}{27} x + \frac{1}{3}

y = - \frac{2}{27} x + \frac{1}{3}

sin^{8} (cos (2 x))

f (x) = sin (x) + x cos (x)

\int 9 sin^{3} (x) cos^{7} (x) d x

\int_{0}^{\infty} x d x

What is the tangent of f(x)= 1/(x^2),(3, 1/9) ?
The tangent of f(x)= 1/(x^2),(3, 1/9) is y=-2/27 x+1/3