Fig wrote:
hobbit wrote:
the reason why checking one point of the region is enough to infer that the whole region share the property is beyond the scope of the discussion here (i can tell you if you really want)...
but this is a good method for solving inequalities in almost all cases.
I must disagree here... One more time, it works on specific question. But, what about the others?
Could u define the scope of this method ?... In these cases, people will use this short cut
actually this method works for virtually any inequality !!!!!!
(the real condition for it to work is that both sides will be continuous functions...for those who know the concept.... for practical uses - it is every inequality that is defined for all numbers)
i'll explain more....
the methods work like this:
to solve an inequality: expression1>expression2, where both expressions are defined for all x.
a) solve the equation: expression1=expression2
b) put the solutions on the number line. this divide it into segments.
c) choose one number for each segment created. for each such value - check the original inequality. if it holds - then the whole segment solves the inequality.
the solution is the union of all segments whose representative "passed" the test in stage c.
the reason this method works is this -
suppose i didn't work. this means that there exists two numbers a and b in the same segment for one exp1>exp2 and for the other exp1<exp2.
but (and this is the complicated part of the method). if it was the case, there was a certain number c between a and b for which exp1=exp2. this is the "continuity assumption" - you can't go from negative to positive without crossing through 0.
but there exists such c between a and b - then it means that a and b are NOT in the same segment. which contradicts our assumption.
so there can't be c like that, and it can't be that exp1<exp2 for a and exp1>exp2 for b for every a,b in the same segment.
so it is sufficient to check one number for each segment to check whether the original inequality holds for the segment.
this method can be extended to work for expressions that have some points where they are undefined.....
and to calm all people down.... most if not all expressions that you'll get into in life are continuous. so the method is quite robust.