What is the tangent of f(x)=((x+3)/(x-3))^2,\at x=4 ?

The tangent of f(x)=((x+3)/(x-3))^2,\at x=4 is y=-84x+385

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tangent of f(x)=((x+3)/(x-3))^2,\at x=4

tangent of

f (x) = (\frac{x + 3}{x - 3})^{2}, at x = 4

y = - 84 x + 385

Solution steps

Find the tangent point:

(4, 49)

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 = 4 : m = - 84

Find the line with slope m=

- 84

and passing through

(4, 49) : y = - 84 x + 385

y = - 84 x + 385

\frac{d}{d x} (ln (\frac{( 1 - x )}{1 + x}))

\int_{1}^{9} π (x - 1)^{2} d x

s \to 0 lim (\frac{3}{s \cdot ( s + 2 )})

\frac{x ^{3}}{9}

(sin (x y) + x y cos (x y)) d x + (x^{2} cos (x y)) d y = 0

What is the tangent of f(x)=((x+3)/(x-3))^2,\at x=4 ?
The tangent of f(x)=((x+3)/(x-3))^2,\at x=4 is y=-84x+385