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

The tangent of f(x)=((x-3)/(x+3))^2,\at x=-2 is y=-60x-95

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

tangent of

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

y = - 60 x - 95

Solution steps

Find the tangent point:

(- 2, 25)

Find the slope of

f (x) = (\frac{x - 3}{x + 3})^{2} : \frac{df ( x )}{d x} = \frac{12 ( x - 3 )}{( x + 3 ) ^{3}}

Compute the slope of

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

x = - 2 : m = - 60

Find the line with slope m=

- 60

and passing through

(- 2, 25) : y = - 60 x - 95

y = - 60 x - 95

y^{^{''}} + 7 y^{^{'}} + 10 y = - 18 t e^{3 t}

\int 4 e^{6 x} d x

\frac{\partial}{\partial x} (cos^{6} (x^{7} y^{8}))

\int_{\frac{1}{2}}^{1} (x^{- 3} - 6) d x

3 x d x + 4 y d y = 0

What is the tangent of f(x)=((x-3)/(x+3))^2,\at x=-2 ?
The tangent of f(x)=((x-3)/(x+3))^2,\at x=-2 is y=-60x-95