What is the tangent of f(x)=(-8)/(x-1) ?

The tangent of f(x)=(-8)/(x-1) is y= 8/((a_{0)-1)^2}x+(-16a_{0}+8)/((a_{0)-1)^2}

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tangent of f(x)=(-8)/(x-1)

tangent of

f (x) = \frac{- 8}{x - 1}

y = \frac{8}{( a _{0} - 1 ) ^{2}} x + \frac{- 16 a _{0} + 8}{( a _{0} - 1 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - \frac{8}{a _{0} - 1})

Find the slope of

f (x) = \frac{- 8}{x - 1} : \frac{df ( x )}{d x} = \frac{8}{( x - 1 ) ^{2}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{8}{( a _{0} - 1 ) ^{2}}

Find the line with slope m=

\frac{8}{( a _{0} - 1 ) ^{2}}

and passing through

(a_{0}, - \frac{8}{a _{0} - 1}) : y = \frac{8}{( a _{0} - 1 ) ^{2}} x + \frac{- 16 a _{0} + 8}{( a _{0} - 1 ) ^{2}}

y = \frac{8}{( a _{0} - 1 ) ^{2}} x + \frac{- 16 a _{0} + 8}{( a _{0} - 1 ) ^{2}}

y^{^{''}} + 2 y^{^{'}} - 3 y = 0, y (0) = 6, y^{^{'}} (0) = - 2

y^{^{''}} + 5 y^{^{'}} = 0

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

f (x) = x^{5}, at x = 1

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

What is the tangent of f(x)=(-8)/(x-1) ?
The tangent of f(x)=(-8)/(x-1) is y= 8/((a_{0)-1)^2}x+(-16a_{0}+8)/((a_{0)-1)^2}