Question #e4a64

1 Answer
Feb 1, 2018

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]}