What is the tangent of f(x)=(x+3)/(x-3) ?

The tangent of f(x)=(x+3)/(x-3) is y=-6/((a_{0)-3)^2}x+(a_{0}+3)/(a_{0)-3}+(6a_{0})/((a_{0)-3)^2}

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tangent of f(x)=(x+3)/(x-3)

tangent of

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

y = - \frac{6}{( a _{0} - 3 ) ^{2}} x + \frac{a _{0} + 3}{a _{0} - 3} + \frac{6 a _{0}}{( a _{0} - 3 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{a _{0} + 3}{a _{0} - 3})

Find the slope of

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

Compute the slope of

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

x = a_{0} : m = - \frac{6}{( a _{0} - 3 ) ^{2}}

Find the line with slope m=

- \frac{6}{( a _{0} - 3 ) ^{2}}

and passing through

(a_{0}, \frac{a _{0} + 3}{a _{0} - 3}) : y = - \frac{6}{( a _{0} - 3 ) ^{2}} x + \frac{a _{0} + 3}{a _{0} - 3} + \frac{6 a _{0}}{( a _{0} - 3 ) ^{2}}

y = - \frac{6}{( a _{0} - 3 ) ^{2}} x + \frac{a _{0} + 3}{a _{0} - 3} + \frac{6 a _{0}}{( a _{0} - 3 ) ^{2}}

\int_{- 1}^{2} 3 x 4 x^{2} - 2 d x

x \to 0 - lim (\frac{1}{e ^{x} - 1})

\int 0.815 x e^{- 0.3 x} d x

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

\frac{d}{d t} (e^{- s t})

What is the tangent of f(x)=(x+3)/(x-3) ?
The tangent of f(x)=(x+3)/(x-3) is y=-6/((a_{0)-3)^2}x+(a_{0}+3)/(a_{0)-3}+(6a_{0})/((a_{0)-3)^2}