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

The tangent of f(x)= 3/(x^2),(1,3) is y=-6x+9

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

tangent of

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

y = - 6 x + 9

Solution steps

Find the slope of

f (x) = \frac{3}{x ^{2}} : \frac{df ( x )}{d x} = - \frac{6}{x ^{3}}

Compute the slope of

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

(1, 3) : m = - 6

Find the line with slope m=

- 6

and passing through

(1, 3) : y = - 6 x + 9

y = - 6 x + 9

\frac{d}{d x} ((9 x - 6)^{2})

\int \frac{1}{10 + 4 x - x ^{2}} d x

\int - \frac{2 x y}{( x ^{2} + y ^{2} ) ^{2}} d y

x y^{^{'}} + 3 y = \frac{( 2 e ^{- 2 x} )}{x ^{2}}

x \to \infty lim (7 + \frac{1}{x})

What is the tangent of f(x)= 3/(x^2),(1,3) ?
The tangent of f(x)= 3/(x^2),(1,3) is y=-6x+9