Triangle Inequality: Given that |x-2|<(1/3), Prove 4 < |3x-11| < 6

[StillAlive]

Help Please!

The Triangle Inequality Theorem states that ||x|-|y||<|x+y|<|x|+|y|

Given that |x-2|<(1/3)
Prove 4 < |3x-11| < 6

The solution for x is 5/3 < x < 7/3. But I cannot just say "as x is approaching" these values since using the proof is required. Anyone has an idea on how to manipulate |3x-11| to fit into the theorem?

PLEASE HELP!!!!! T.T

 

[ksdhart2]

Well, I see two options you can employ. For the first option, I'll use the notation f(x)=|3x-11| for ease. Option one is to prove that f(x) must always be between 4 and 6, but never reach either of those values. You can graph the portion of that f(x) between x=5/3 and x=7/3 and see that it is the inequality holds and that the function is monotonically decreasing (i.e. it only ever decreases, never increases). Furthermore, you can see that at x=5/3, f(x)=6 and at x=7/3, f(x)=4. Now, if you could prove that the function is monotonic on that interval, you'd have proven that f(x) must always lie between 4 and 6. If f(x) is decreasing, what then can you say about its derivative? Can you see how that helps you in the proof?

Option two might be a bit easier, if you prefer this instead. You want to manipulate the given inequality such that it is of the same form as the "triangle inequality theorem." The middle portion, i.e. |3x-11| needs to be of the form |x+y|. Well, if we note that |3x-11| = 3|x-11/3|, where does this line of reasoning take you? What must the value of y be? What if you divide all three inequalities by 3? Try playing with this and see where it leads you.