Proving GLB of polynomial

[mdrak12]

given set B = {x^2+2x-3} where x is a real number

I needed to prove the existence of the infimum. The definition I used was "Greatest lower bound property tells us that every non-empty set of real numbers which is bounded from below has an infimum."

I know the correct answer is -4. I proved this through finding a critical point and then using the first derivative test to show it is a minimum.

I then went on to justify why -4 is the greatest lower bound by saying:

1) To prove that -4 is the greatest lower bound, we will prove that -4 is the absolute minimum. That is because, by definition the absolute minimum is the smallest value that the function can have over its entire curve.

2) The function g(x) is a second-degree polynomial where the leading coefficient is positive. This means that the graph is a parabola and is concave up, with no endpoints.

3) And so, g(x) approaches infinity as x goes to positive infinity. g(x) approaches infinity as x goes to negative infinity.

4) Thus, g(x) is greater than or equal to -4, for all x, where x is a real number.

I got the feedback from my evaluater saying,
"it was not evident how this work related to the definition of infimum or to the intended approach using "every non-empty set of real numbers which is bounded from below has an infimum".

What else do I need to add to my justification?

 

[tkhunny]

1) "goes to infinity" is a silly expression. Feel free to get it out of your vocabulary.
2) Does an infimum have to be IN the bounded set?
3) "The function ... is a ... polynomial." Is that true?
4) What if I bring you a number, M, from your set, and tell you it's less than -4? Prove me wrong.
5) I'm telling you that $$m+\epsilon$$ is the lowest bound on your set. Prove to me that there is something less than that.

 

[Romsek]

What you don't really address is the question of whether any number greater than -4 can be a lower bound of the function.

It's pretty obvious that it can't be, which is probably why you ignored it, but it's a critical step in showing a value is a GLB

 

[pka]

given set B = {x^2+2x-3} where x is a real number
I needed to prove the existence of the infimum. The definition I used was "Greatest lower bound property tells us that every non-empty set of real numbers which is bounded from below has an infimum."
I know the correct answer is -4. I proved this through finding a critical point and then using the first derivative test to show it is a minimum.

I hope that the question means that $\displaystyle B=\{(x,y)=(x,x^2+2x-3): x\in\mathbb{R}\}$
That is a parabola that opens up , therefore its vertex in the lowest point on the graph.
If it is asking about a bound for the y-values then O.K.. Otherwise the question is meaningless.