Applying function to sides of a disequality

A disequation like this

-1 <= sinx <= 1

of course is always true, and is equivalent to

-1 <= sinx

sinx <= 1

What if you apply f(x) = x^-1 to both sides? It's a monotically decreasing function for x != 0, so you should just

reverse the disequality... but!

-1 >= 1/sinx

1 <= 1/sinx

which is nonsense. Am I wrong somewhere?

Re: Applying function to sides of a disequality

It's probably not total nonsense. There are issues arising because of the asymptotes but at least 1/sinx never gives a value BETWEEN -1 and 1 as the graph shows.

plot(1/(sinx)) - Wolfram|Alpha

:)

Note*

is equivalent to

and in interval notation

The INTERSECTION of the 2 intervals

but the other one is OR if you interpret it as AND there is no solution.

OR

and in interval notation

The UNION of the 2 intervals.

:)

Re: Applying function to sides of a disequality

Thanks a lot!

But then, a chained disequality is an and-equivalent. Why applying a function should change that?

Re: Applying function to sides of a disequality

Quote:

Originally Posted by

**AndrewTahoe** Thanks a lot!

But then, a chained disequality is an and-equivalent. Why applying a function should change that?

I edited my post to include interval notation but if that doesn't help with your new question consider...

Now apply the inverse to everything

What can we put for the question marks ??

we see that

Will help us with the dilemmas of -??-

:)

Re: Applying function to sides of a disequality

Hi,

A real valued function on a set D of reals is decreasing provided for all a and b __in__ D with a < b, f(a) > f(b). You said f(x)=1/x is decreasing everywhere but 0. In a sense this is right. That is, f is decreasing on and decreasing on . But this does not mean f is decreasing on . From -2 < 1 it certainly does not follow that -1/2 > 1/1.

So for your original inequality, if sin(x) > 0 then 1 and sin(x) are in a set D where f(x)=1/x is decreasing and so . But if sin(x) < 0, you can't conclude .

Re: Applying function to sides of a disequality

Ok, that was clear. In fact, the definition of monotonically decreasing make sense when we consider an interval that is a subset of the domain of the function. No "holes" allowed!

Thanks!