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

The tangent of f(x)= x/(x-2),\at x=3 is y=-2x+9

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

tangent of

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

y = - 2 x + 9

Solution steps

Find the tangent point:

(3, 3)

Find the slope of

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

Compute the slope of

f (x) = \frac{x}{x - 2}

x = 3 : m = - 2

Find the line with slope m=

- 2

and passing through

(3, 3) : y = - 2 x + 9

y = - 2 x + 9

(- 5.2) (2.5)

\frac{d}{d x} (sin (x + x) + 5)

\frac{\partial}{\partial x} (6 + x ln (x y - 9))

f (x) = x^{2} - 3 x + 2

\frac{s + 1}{( s - 1 ) \cdot ( s - 2 )}

What is the tangent of f(x)= x/(x-2),\at x=3 ?
The tangent of f(x)= x/(x-2),\at x=3 is y=-2x+9