What is the tangent of f(x)= 1/((x-1)^2),\at x=0 ?

The tangent of f(x)= 1/((x-1)^2),\at x=0 is y=2x+1

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tangent of f(x)= 1/((x-1)^2),\at x=0

tangent of

f (x) = \frac{1}{( x - 1 ) ^{2}}, at x = 0

y = 2 x + 1

Solution steps

Find the tangent point:

(0, 1)

Find the slope of

f (x) = \frac{1}{( x - 1 ) ^{2}} : \frac{df ( x )}{d x} = - \frac{2}{( x - 1 ) ^{3}}

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 = 2

Find the line with slope m=

2

and passing through

(0, 1) : y = 2 x + 1

y = 2 x + 1

\int 2 π x d x

y = x^{2} - 3

y^{^{'}} e^{- x} = y^{2} + 9

\frac{d ^{2}}{d x ^{2}} (2 cos (\frac{x}{2}))

2 e^{x^{2} - 5 x}

What is the tangent of f(x)= 1/((x-1)^2),\at x=0 ?
The tangent of f(x)= 1/((x-1)^2),\at x=0 is y=2x+1