Inequalities and direction-flipping

October 19, 2007 10:00 PM

Why is it that when I solve the inequality 0 ≤ arccos(x) <> cos(π/4)? Why do the operators switch direction?

I understand that when multiplying or dividing by a negative number, the operators switch direction - but why is this the case for functions?

posted by PuGZ to education (11 comments total)

Ugh, I have no idea what happened there. The preview was fine. My question was:

Why is it that when I solve the inequality 0 ? arccos(x) < ?/4, the answer is cos(0) ? x > cos(?/4)? Why do the operators switch direction?
posted by PuGZ at 10:02 PM on October 19, 2007

..And now my less than or equal to ( =<> ) don't render in that comment.
posted by PuGZ at 10:04 PM on October 19, 2007

Less than or equal to: =<
Pi: 3.14159..
Greater than or equal to: =>

Today is not my day.
posted by PuGZ at 10:05 PM on October 19, 2007

arccos is always going to be greater or equal to 0.

I don't understand your question though. Are you trying to solve this inequality: here?
posted by demiurge at 10:51 PM on October 19, 2007

I'm trying to solve this inequality. AskMefi seems to have chewed up the rendering of my question, though.
posted by PuGZ at 11:18 PM on October 19, 2007

So, you're solving it by applying cos to both sides?

cos is a monotonically decreasing function (at least in the range you're considering), so applying it to both sides requires you to flip the sign.
posted by demiurge at 11:34 PM on October 19, 2007 [1 favorite]

So I take it that if the functions in question were sin or tan, the sign needn't be flipped. My (high-school) textbook uses cos as an example and doesn't explain why the sign was flipped.

Thankyou very much for putting it so clearly. :-)
posted by PuGZ at 11:44 PM on October 19, 2007

Just to throw this in: the function x -> ax ("multiply by a") is monotone decreasing when a is negative, monotone increasing when a is positive. So the fact that you have to flip signs when multiplying both sides by a negative a is another example of what demiurge points out -- applying a monotone decreasing function to both sides induces a sign flip.

Exercise: what happens if you apply a non-monotone function to each side?
posted by escabeche at 5:48 AM on October 20, 2007

This is nothing more than blind guessing, but would it depend on the instantaneous rate of change at a given point?

Had the points been 0 and 5/4 * pi rather than 0 and pi/4, would the sign only change for the cos(0)?
posted by PuGZ at 6:54 AM on October 20, 2007

In case of functions that go down then up, like absolute value, you have to split the inequality into cases. I'm not sure what you have to do if you have even more screwy functions, like something that can't be broken down into a sequence of monotonically increasing and monotonically decreasing parts.
posted by demiurge at 9:23 AM on October 20, 2007

Multiplying by -1 is the simplest example of a decreasing function:

1 < 10
-1 > -10

(You can think of this as applying f(x) = -1*x to both sides.)
posted by mbrubeck at 8:33 PM on October 20, 2007