prove $f + g ∈ Θ(max(f, g))$

Question

I want to prove $f + g ∈ Θ(max(f, g))$

So I did that $f,g \geq 0$, then $\frac{1}{2}(f+g) \leq \max(f,g) \leq (f+g)$ is equivalent to $ \max(f,g) \leq (f+g) \leq 2 \max(f,g)$.

What if $f \leq 0$ or $g \leq 0$, what can I conclude about $(f + g)$'s time complexity? It seems to be not determinable.

Answer 1

You are right. If $f \ge 0$ and $g \le 0$ you cannot get this conclusion. (Or any conclusion.) In fact, you could have $g=-f$ and get $f+g=0$ regardless of how big $f$ is.