Proving that $f(x)=2^x$ is $O(x^2)$ Can someone help me with this problem? I don't really know what to do if the x is in exponential form. 

A direct proof (that is, not by contradiction):
For every $z\geqslant0$, one has $2^z\gt z$. Using this for $z=x/3$ and raising both sides to their cubes yields $2^x\gt x^3/27$ for every $x\geqslant0$. In particular, $2^x/x^2\geqslant x/27$ hence $2^x/x^2\to+\infty$ when $x\to+\infty$.
To prove that $2^z\gt z$ for $z\gt1$, one can write $2^z=(1+1)^z\gt1+z\cdot1\gt z$.
