What is the tangent of f(x)=((x+2)/(x-2))^2,\at x=6 ?

The tangent of f(x)=((x+2)/(x-2))^2,\at x=6 is y=-x+10

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tangent of f(x)=((x+2)/(x-2))^2,\at x=6

tangent of

f (x) = (\frac{x + 2}{x - 2})^{2}, at x = 6

y = - x + 10

Solution steps

Find the tangent point:

(6, 4)

Find the slope of

f (x) = (\frac{x + 2}{x - 2})^{2} : \frac{df ( x )}{d x} = - \frac{8 ( x + 2 )}{( x - 2 ) ^{3}}

Compute the slope of

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

x = 6 : m = - 1

Find the line with slope m=

- 1

and passing through

(6, 4) : y = - x + 10

y = - x + 10

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

\frac{d}{d x} (2 x^{\frac{1}{4}} (x^{3} - 11))

f (x) = 5 x^{2} + 4

\frac{d}{d x} (\frac{- 2}{2 x + 5})

\frac{\partial}{\partial y} ((x + y) (x^{2} + 2 y^{2}))

What is the tangent of f(x)=((x+2)/(x-2))^2,\at x=6 ?
The tangent of f(x)=((x+2)/(x-2))^2,\at x=6 is y=-x+10