# Thread: Applying function to sides of a disequality

1. ## 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?

2. ## 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*

$-1 \le sinx \le 1$

is equivalent to

$-1 \le sinx \ \ AND \ \ sinx \le 1$

and in interval notation

$[-1, \infty) \cap (- \infty, 1] = [-1,1]$

The INTERSECTION of the 2 intervals

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

$-1 \ge \frac{1}{sinx}$

OR

$1 \le \frac{1}{sinx}$

and in interval notation

$(- \infty , -1] \cup [1 , \infty)$

The UNION of the 2 intervals.

3. ## 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?

4. ## Re: Applying function to sides of a disequality

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...

$-2 \le 2 \le 4$

Now apply the inverse to everything

$-\frac{1}{2} ?? \frac{1}{2} ?? \frac{1}{4 }$

What can we put for the question marks ??

we see that

$-\frac{1}{2} \le \frac{1}{2} \ \ OR \ \ \frac{1}{2} \ge \frac{1}{4}$

Will help us with the dilemmas of -??-

5. ## 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 $(-\infty,0)$ and decreasing on $(0,\infty)$. But this does not mean f is decreasing on $(-\infty,0)\cup(0,\infty)$. 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 $1\geq 1/sin(x)$. But if sin(x) < 0, you can't conclude $1/sin(x)\geq1$.

6. ## 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!