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

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

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

tangent of

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

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

Solution steps

Find the tangent point:

(2, 1)

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}

x = 2 : 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}

\int sin^{2} (θ) cos^{5} (θ) d θ

\frac{\partial}{\partial x} (e^{5 x y})

\frac{d}{d x} (sec^{2} (x) tan^{2} (x))

\int \frac{2 x + 4}{x ^{2} - x} d x

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