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

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

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tangent of 1/((x-2))

tangent of

\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

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

Compute the slope of

y = \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}}

\frac{d ^{2} y}{d t ^{2}} + 3 \frac{d y}{d t} - 40 y = 1 - 4 t

\int_{2}^{\infty} x e^{- 8 x} d x

e^{x}, x e^{x}, 0, 1

y = \frac{x ^{4}}{4} + \frac{1}{8 x ^{2}}

\int - \frac{1}{2} cos (2 x) d x

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