What is the tangent of f(x)= 2/(x+1),\at x=3,y=1 ?

The tangent of f(x)= 2/(x+1),\at x=3,y=1 is y=-1/8 x+7/8

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tangent of f(x)= 2/(x+1),\at x=3,y=1

tangent of

f (x) = \frac{2}{x + 1}, at x = 3, y = 1

y = - \frac{1}{8} x + \frac{7}{8}

Solution steps

Find the tangent point:

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

Find the slope of

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

Compute the slope of

f (x) = \frac{2}{x + 1}

x = 3 : m = - \frac{1}{8}

Find the line with slope m=

- \frac{1}{8}

and passing through

(3, \frac{1}{2}) : y = - \frac{1}{8} x + \frac{7}{8}

y = - \frac{1}{8} x + \frac{7}{8}

x - 2 x

\int \frac{e ^{- x}}{x} d x

3 x y + y^{2} + (x^{2} + x y) y^{^{'}} = 0

\frac{d y}{d x} = \frac{x ^{5} - 4 y}{x}

\int \frac{x}{4 - x ^{4}} d x

What is the tangent of f(x)= 2/(x+1),\at x=3,y=1 ?
The tangent of f(x)= 2/(x+1),\at x=3,y=1 is y=-1/8 x+7/8