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}}

Compute the slope of

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

x = 0 : m = - 2

Find the line with slope m=

- 2

and passing through

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

y = - 2 x - 1

5 t \cdot \frac{d y}{d t} + 4 y = t

\int_{2}^{\infty} \frac{2}{t ^{2} - 1} d t

(9 - x^{2}) (\frac{1}{3}) π x^{2}

\int_{0}^{16} \frac{1}{16 - x} d x

f (x) = (- 3 x^{2} + 3) e^{2 x}, at x = 2

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