When solving inequalities are we more concerned about the denominator being undefined or with the numerator being zero.

For example in the question [\frac{(2x+1)^2(x-1)}{x(x^2-1)}/=0]

The solution is (-infinity, 1)U{-1/2}U(0,1)U(1,infinity)

And I was wondering if a box bracket should be on the one... but then I noticed it is in the denominator and would make the inequality undefined so... here raises the question once again.

Question

When solving inequalities are we more concerned about the denominator being undefined or with the numerator being zero.

For example in the question [\frac{(2x+1)^2(x-1)}{x(x^2-1)}>/=0]

The solution is (-infinity, 1)U{-1/2}U(0,1)U(1,infinity)

And I was wondering if a box bracket should be on the one... but then I noticed it is in the denominator and would make the inequality undefined so... here raises the question once again.

anonymous · Answer

$$\frac{(2x+1)^2(x-1)}{x(x^2-1)}\ge0$$