At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
-y < f(x) < y
|f(x)| > y -> f(x) > y , f(x) < -y
can you do it now ?
um, no, I don't get what you mean by all that..
ok if we divide the equation by -1 what will we get ?
well, my teacher said to change the < to a = when I solve these types of problems so i would get: |2m-1| = -2
ok if you want to do it this way you dont have to divide.. but if you done it already now you know to find m's that satisfy |2m-1| = -2 ?
well, to get rid of the absolute values first, I just seperated it into 2m -1 = -2 and 2m -1 = 2 like my teacher told me to and I got the answers -1/2 and 3/2 but when I tried to check it, it doesn't seem right..
you have to plug now some m<-1/2 like m = -1 -1/2 < m < 3/2 lke m = 1 and m>3/2 like m = 2 in the original equation and see if it holds
when does it hold?
are you there?
yeah sorry, I tried numbers but it didn't work...
can someone else tell me how to solve this or if I'm doing something wrong?
why? when you plug m=-1 what do you get?
i get -4 <2
so it holds isnt it?
-4 is indeed smaller than 2
yes, I know but the answers i got for m and every other number works too which shouldn't be happening
it works for every m
it is obvious.. the original question is negative number < 2 so it always holds i wanted to work on the technique with you
i don't get how that can work though...so the answer is that all numbers work?
yes it is obvious .. all the negative numbers are smaller than 2
if you didnt see it at the original question you could see it when you asked yourself |2m-1| = -2 ? it is never happens .. since left hand side of the equation is always positive !!
ohhhh, okay, thank you :)