# Is #f(x)=sqrt(1/x-2) # increasing or decreasing at #x=2 /9 #?

Look at the function or take the derivative to find that

Applying the power rule, the chain rule, and the quotient rule gives us

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To determine if the function ( f(x) = \sqrt{\frac{1}{x} - 2} ) is increasing or decreasing at ( x = \frac{2}{9} ), we can examine the sign of its derivative at that point.

First, find the derivative of ( f(x) ) with respect to ( x ): [ f'(x) = -\frac{1}{2x^2\sqrt{\frac{1}{x} - 2}} ]

Next, evaluate the derivative at ( x = \frac{2}{9} ): [ f'\left(\frac{2}{9}\right) = -\frac{1}{2\left(\frac{2}{9}\right)^2\sqrt{\frac{1}{\frac{2}{9}} - 2}} ] [ f'\left(\frac{2}{9}\right) = -\frac{1}{2\left(\frac{2}{81}\right)\sqrt{9 - 2}} ] [ f'\left(\frac{2}{9}\right) = -\frac{1}{2\left(\frac{2}{81}\right)\sqrt{7}} ] [ f'\left(\frac{2}{9}\right) = -\frac{81}{4\sqrt{7}} ]

Since the derivative is negative at ( x = \frac{2}{9} ), the function is decreasing at that point.

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