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

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

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

tangent of

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

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

Solution steps

Find the tangent point:

(1, - \frac{1}{2})

Find the slope of

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

Compute the slope of

f (x) = \frac{x}{x - 3}

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

Find the line with slope m=

- \frac{3}{4}

and passing through

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

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

\frac{\partial}{\partial x} (3 x^{2} + x y + y^{2} - 8)

\frac{\partial}{\partial x} (\frac{6}{x - 2} + 7 x)

\int (x^{2} - 5 x + 2) d x

\frac{d}{d x} (lo g_{e} (3 x))

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

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