What is the tangent of f(x)= 1/(x-2) ?

The tangent of f(x)= 1/(x-2) is y=-1/((a_{0)-2)^2}x+(2a_{0}-2)/((a_{0)-2)^2}

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tangent of f(x)= 1/(x-2)

tangent of

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

y = - \frac{1}{( a _{0} - 2 ) ^{2}} x + \frac{2 a _{0} - 2}{( a _{0} - 2 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{a _{0} - 2})

Find the slope of

f (x) = \frac{1}{x - 2} : \frac{df ( x )}{d x} = - \frac{1}{( x - 2 ) ^{2}}

Compute the slope of

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

x = a_{0} : m = - \frac{1}{( a _{0} - 2 ) ^{2}}

Find the line with slope m=

- \frac{1}{( a _{0} - 2 ) ^{2}}

and passing through

(a_{0}, \frac{1}{a _{0} - 2}) : y = - \frac{1}{( a _{0} - 2 ) ^{2}} x + \frac{2 a _{0} - 2}{( a _{0} - 2 ) ^{2}}

y = - \frac{1}{( a _{0} - 2 ) ^{2}} x + \frac{2 a _{0} - 2}{( a _{0} - 2 ) ^{2}}

y^{^{'}} - 2 x (y^{2} + 1) = 0

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

x \to 0 lim (\frac{x + 1}{x + 4})

(\frac{d y}{d x}) = \frac{( 2 x ( y - 1 ) )}{x ^{2} + 3}

What is the tangent of f(x)= 1/(x-2) ?
The tangent of f(x)= 1/(x-2) is y=-1/((a_{0)-2)^2}x+(2a_{0}-2)/((a_{0)-2)^2}