tertre

I do AoPs, and we get a lot of problems, but it helps!

Thank you, CPhill! 

85^2-83^2=336, so it is true..

Or, (a+b)(a-b)=(85+83)(85-83)=168*2=336.

??? I'll check my work again, are they asking for the 85 or 7225...

We have \(x\) as the smaller odd number and \(x+2\) as the larger odd number.

Translating this into an equation, we have \((x+2)^2-x^2=336\).

Expanding \((x+2)^2-x^2=4x+4\).

Rewriting, we attain \(4x+4=336, 4x=332\) and \(x=83.\)

So,  \(x+2=85\) or \(\boxed{7225}.\)

Yes, you are correct CPhill. Misread the problem!

Hint: The numbers when closer together have the maximum area.

Factoring:: \(-16t^2+80t+21=t=-\frac{1}{4},\:t=\frac{21}{4}\)

We first factor \(-16t^2+80t+21\) into \(-\left(4t+1\right)\left(4t-21\right)\)

Next, \(4t+1=0 \longrightarrow t=-\frac{1}{4}.\)

And, \(4t-21=0 \longrightarrow t=\frac{21}{4}.\)

Since the value has to be maximum, the answer is \(\boxed{\frac{21}{4}}.\)

Oh, I think I'm wrong....

Yes, you're correct! Since we have \(f(x)=f(2-x)\), \(2x=2, x=1.\)

I just read about pretty hard functions, hope I can solve it!!

\(g(x)=f(f(x))=\frac{1}{2x-5}=\frac{1}{2(\frac{1}{2x-5})-5}=\frac{2x-5}{27-10x}\)

Thus, the largest \(x\) not in the domain of \(g(x)\) is .\(\boxed{\frac{27}{10}}.\)