What is the tangent of f(x)= 1/(x-2),\at x=6 ?

The tangent of f(x)= 1/(x-2),\at x=6 is y=-1/16 x+5/8

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tangent of f(x)= 1/(x-2),\at x=6

tangent of

f (x) = \frac{1}{x - 2}, at x = 6

y = - \frac{1}{16} x + \frac{5}{8}

Solution steps

Find the tangent point:

(6, \frac{1}{4})

Find the slope of

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

Compute the slope of

f (x) = \frac{1}{x - 2}

x = 6 : m = - \frac{1}{16}

Find the line with slope m=

- \frac{1}{16}

and passing through

(6, \frac{1}{4}) : y = - \frac{1}{16} x + \frac{5}{8}

y = - \frac{1}{16} x + \frac{5}{8}

\int \frac{1}{2 sin ( x ) - cos ( x ) + 5} d x

\frac{d}{d x} ((5)^{x})

\frac{\partial}{\partial x} (x^{n})

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

What is the tangent of f(x)= 1/(x-2),\at x=6 ?
The tangent of f(x)= 1/(x-2),\at x=6 is y=-1/16 x+5/8