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

The tangent of f(x)=(3x)/(x+1),\at x=2 is y= 1/3 x+4/3

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

tangent of

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

y = \frac{1}{3} x + \frac{4}{3}

Solution steps

Find the tangent point:

(2, 2)

Find the slope of

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

Compute the slope of

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

x = 2 : m = \frac{1}{3}

Find the line with slope m=

\frac{1}{3}

and passing through

(2, 2) : y = \frac{1}{3} x + \frac{4}{3}

y = \frac{1}{3} x + \frac{4}{3}

\frac{4 x}{x ^{2} + 1}

x^{3} y^{^{''}} - x y^{^{'}} + y = 0

\frac{d}{d x} (\frac{( x + a ) ^{3}}{b} - c)

\int \frac{tan ( x )}{sin ( 2 x )} d x

\int \frac{- x}{x + 1 - x + 1} d x

What is the tangent of f(x)=(3x)/(x+1),\at x=2 ?
The tangent of f(x)=(3x)/(x+1),\at x=2 is y= 1/3 x+4/3