Is the Function f(x) = (x+2)^-1 Bounded on the Open Interval (-2,2)?

[Benny]

Hi, I would like to know if the function f(x) = (x+2)^-1 is bounded on the open interval (-2,2)? The interval doesn't include the point x = -2 but I'm not sure if I can say that there is a K>=0 such that |f(x)| < K for all x in (-2,2).

The function is defined everywhere in that interval but still approaches positive infinity as x tends to x = - 2 from above so I'm not sure what to conclude here. Any help would be great thanks.

 

[quasar987]

Instead of trying to find such a K, i.e. instead of trying to proove that f is bounded, try to prove that it is not bounded. (And it obviously is not bounded since, as you noted yourself, it is intuitively evident that the function goes to infinity as x gets nearer and nearer to -2)

A function is not bounded on its domain if given any number M, you can find an element 'a' of the domain such that f(a)>M.

So somewhat like in the epsilon-delta proofs, if you can find a relation a(M) that associated to every M a number 'a' in (-2,2) such that f(a)>M, then you will have proven the unboundedness of f.

 

[JasonRox]

Do you know anything about finding the maximum of a function using derivatives?

If yes, find the derivative of the function and try to find a maximum/minimum point. You shouldn't be able to find one, which means the function itself isn't bounded in the open interval.

Otherwise, assume there exists some K that does satisfy your equation. Now, find an x that does not satisfy the equation, which is a contradiction so that such a K can not exist.

Therefore, simply solve |f(x)|<K, where K is a constant and be sure to pick an x that is in the open interval (-2,2).

Note: The answer for x should be dependent on K.

 

[Benny]

Ok thanks. I probably made it more complicated than it was.