What is the tangent of f(x)=(6x)/(x+2),\at x=4 ?

The tangent of f(x)=(6x)/(x+2),\at x=4 is y= 1/3 x+8/3

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tangent of f(x)=(6x)/(x+2),\at x=4

tangent of

f (x) = \frac{6 x}{x + 2}, at x = 4

y = \frac{1}{3} x + \frac{8}{3}

Solution steps

Find the tangent point:

(4, 4)

Find the slope of

f (x) = \frac{6 x}{x + 2} : \frac{df ( x )}{d x} = \frac{12}{( x + 2 ) ^{2}}

Compute the slope of

f (x) = \frac{6 x}{x + 2}

x = 4 : m = \frac{1}{3}

Find the line with slope m=

\frac{1}{3}

and passing through

(4, 4) : y = \frac{1}{3} x + \frac{8}{3}

y = \frac{1}{3} x + \frac{8}{3}

y^{^{''}} - 4 y^{^{'}} + 4 y = e^{2 x}

(5 y - 2 x) d y - 2 y d x = 0

\frac{\partial}{\partial x} (\frac{3 x - 5 y}{3 x + 5 y})

\int sin (8 x) d x

\frac{\partial}{\partial y} (e^{x y} + 2 x - 5 y)

What is the tangent of f(x)=(6x)/(x+2),\at x=4 ?
The tangent of f(x)=(6x)/(x+2),\at x=4 is y= 1/3 x+8/3