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

Dec 5, 2015

Look at the function or take the derivative to find that $f \left(x\right)$ is decreasing.

#### Explanation:

If we think about what $f \left(x\right)$ is actually doing, we can see that it is defined on $\left(0 , \frac{1}{2}\right]$, with $f \left(x\right) \to \infty$ as $x \to {0}^{+}$ and $f \left(\frac{1}{2}\right) = 0$.
Then, as $x$ increases, $\frac{1}{x}$ decreases, so $f \left(x\right)$ decreases, meaning $f \left(x\right)$ is decreasing on all points where it is defined.

So we know the answer without doing any calculus, but let's do it the calculus way as well. A function is increasing or decreasing at a certain point if its first derivative is positive or negative, respectively. Then, to confirm our answer of decreasing, the first derivative should be negative at $x = \frac{2}{9}$

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

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} {\left(\frac{1}{x} - 2\right)}^{\frac{1}{2}}$
$= \frac{1}{2} {\left(\frac{1}{x} - 2\right)}^{- \frac{1}{2}} \cdot \left(- \frac{1}{x} ^ 2\right)$
$= - \frac{1}{2 {x}^{2} {\left(\frac{1}{x} - 2\right)}^{\frac{1}{2}}}$

As the denominator is always positive, then we have $f ' \left(x\right) < 0$ for all $x \in \left(0 , \frac{1}{2}\right]$, meaning $f \left(x\right)$ is decreasing throughout the interval, including at $x = \frac{2}{9}$ (note that this matches our analysis above).