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

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

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

tangent of

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

y = 2 x - 1

Solution steps

Find the tangent point:

(0, - 1)

Find the slope of

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

Find the line with slope m=

2

and passing through

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

y = 2 x - 1

t an g e n t x^{2} + x y - y^{2} = - 4, (2, 4)

y^{^{'}} = \frac{5 x ^{5} e ^{\frac{y}{x}} + x ^{3} y ^{2}}{x ^{4} y}

\int (\frac{x + 5}{x ^{2} + 5 x + 6}) d x

in v erse l a pl a ce \frac{1}{s + \frac{1}{2}}

t an g e n t f (x) = 20 (x - 1) e^{- 0.5 x}, at x = 4

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