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

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

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

tangent of

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

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

Solution steps

Find the tangent point:

(3, \frac{1}{9})

Find the slope of

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

\int \frac{5 x ^{3} + 5 x + 11}{( x ^{2} + 1 ) ^{2}} d x

y^{^{'}} - \frac{y}{x} = x^{2}

\frac{2 s + 5}{s ^{2} - 4 s - 5}

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\int sin (x) cos (6 x) d x

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