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

The tangent of f(x)= 3/(x+1),(2,1) is y=-1/3 x+5/3

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

tangent of

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

y = - \frac{1}{3} x + \frac{5}{3}

Solution steps

Find the slope of

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

Compute the slope of

f (x) = \frac{3}{x + 1}

(2, 1) : m = - \frac{1}{3}

Find the line with slope m=

- \frac{1}{3}

and passing through

(2, 1) : y = - \frac{1}{3} x + \frac{5}{3}

y = - \frac{1}{3} x + \frac{5}{3}

y^{^{'}} = e^{- 3 x} + 2 x

\frac{d}{d x} (x (x^{2} + 1)^{10})

\int - \frac{1}{Ux} + \frac{1}{U ( x - U )} d x

\frac{\partial}{\partial y} (\frac{e ^{y}}{x + y ^{2}})

What is the tangent of f(x)= 3/(x+1),(2,1) ?
The tangent of f(x)= 3/(x+1),(2,1) is y=-1/3 x+5/3