Question

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

Answer 1

Look at the function or take the derivative to find that `f(x)` is decreasing.

Explanation:

If we think about what `f(x)` is actually doing, we can see that it is defined on `(0, 1/2]`, with `f(x) -> oo` as `x->0^+` and `f(1/2) = 0`.
Then, as `x` increases, `1/x` decreases, so `f(x)` decreases, meaning `f(x)` 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=2/9`

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

`f'(x) = d/dx(1/x-2)^(1/2)`
`= 1/2(1/x-2)^(-1/2)*(-1/x^2)`
`=-1/(2x^2(1/x-2)^(1/2))`

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