What is the tangent of f(x)= 2/(x-4) ?

The tangent of f(x)= 2/(x-4) is y=-2/((a_{0)-4)^2}x+(4a_{0}-8)/((a_{0)-4)^2}

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tangent of f(x)= 2/(x-4)

tangent of

f (x) = \frac{2}{x - 4}

y = - \frac{2}{( a _{0} - 4 ) ^{2}} x + \frac{4 a _{0} - 8}{( a _{0} - 4 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{2}{a _{0} - 4})

Find the slope of

f (x) = \frac{2}{x - 4} : \frac{df ( x )}{d x} = - \frac{2}{( x - 4 ) ^{2}}

Compute the slope of

f (x) = \frac{2}{x - 4}

x = a_{0} : m = - \frac{2}{( a _{0} - 4 ) ^{2}}

Find the line with slope m=

- \frac{2}{( a _{0} - 4 ) ^{2}}

and passing through

(a_{0}, \frac{2}{a _{0} - 4}) : y = - \frac{2}{( a _{0} - 4 ) ^{2}} x + \frac{4 a _{0} - 8}{( a _{0} - 4 ) ^{2}}

y = - \frac{2}{( a _{0} - 4 ) ^{2}} x + \frac{4 a _{0} - 8}{( a _{0} - 4 ) ^{2}}

\int \frac{4 x ^{5} - 2 x + 3}{x ^{3}} d x

e^{- 2 s} (\frac{1}{s - 9})

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

\frac{d}{d t} (e^{t} \cdot cos (t))

(\frac{1}{2}, 2), (6, 2)

What is the tangent of f(x)= 2/(x-4) ?
The tangent of f(x)= 2/(x-4) is y=-2/((a_{0)-4)^2}x+(4a_{0}-8)/((a_{0)-4)^2}