Question

Question #e4a64

Answer 1

The derivative remains negative over the specified domain, therefore we can conclude that the function is monotonic on the domain provided.

Explanation:

A monotonic function is a function which is either entirely increasing or decreasing. I like the definitions that its first derivative (which need not be continuous) does not change sign.

For the function at hand, it means that the sign of the derivative of `f(x)` must not change sign on the interval `x \in (-1,0)`. First we find the derivative:

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

First, we discover that the derivative only contains `x` squared so the fact that the domain is entirely negative doesn't seem to matter. Next we look at the derivative at `x=-1+\delta` since the end point is not actually included, where `\delta` is positive:

`d/dx f(-1+\delta) = (-2\delta + delta^2)/(1-2\delta + \delta^2)`

From this we can see that for positive `delta` much less than `1`, the numerator goes to `-2\delta` and the denominator goes to `1`. Even in the limit, the derivative approaches `0` but never crosses into positive:

`lim_(delta -> 0^+) (-2\delta + delta^2)/(1-2\delta + \delta^2) =0`

At the other end of the range, where `x -> 0^-`, the derivative approaches looks like:

`lim_(x -> 0^-)-1/x^2 = -infty`

The derivative remains negative over the specified domain, therefore we can conclude that the function is monotonic on the domain provided. A quick graph confirms our conclusion.

graph{1-1/x^2 [-1, 0, -100, 5]}