- Thread starter
- #1

- Thread starter jacks
- Start date

- Thread starter
- #1

- Mar 22, 2013

- 573

The RHS grows superexponentially whether the LHS is quadratic. So, for obvious reasons, they intersects each other at most finitely often. A more close inspection of the behaviour would lead to the fact that they do indeed intersect ones.

PS I have no formal proof of this piece of problem, unfortunately.

Balarka

.

- Moderator
- #3

- Feb 7, 2012

- 2,785

That argument looks correct to me. As $x$ goes from $-\infty $ to $0$, the LHS increases from $0$ to $12$, and the RHS decreases from $\infty$ to $0$. So the graphs must cross exactly once. For $x>0$ the LHS increases (very fast!) from $12$ to $\infty$ and the RHS increases much more slowly. If you differentiate both functions I am sure you will find that the LHS increases faster than the RHS for all positive $x$.The RHS grows superexponentially whether the LHS is quadratic. So, for obvious reasons, they intersects each other at most finitely often. A more close inspection of the behaviour would lead to the fact that they do indeed intersect ones.

PS I have no formal proof of this piece of problem, unfortunately.

- Aug 18, 2013

- 76

Hint:

- Mar 22, 2013

- 573

That is evident, as I mentioned before, since LHS grows superexponentially, i.e., \(\displaystyle << 5^x\) whereas the RHS is quadratic. I think this is sufficient to prove the fact.Opalg said:I am sure you will find that the LHS increases faster than the RHS for all positive x

PS I see eddybob just posted a rigourus argument of the fact here