# Determine whether the function $f(x)$ is of order $2^x$

Prove $$f(x) = 2^x + x^2$$ It is of the order $$2^x$$ or $$f(x)$$ is of the order O($$2^x$$). So far I've got,

$$|f(x)| = 2^x + x^2 \le |2^x| + |x^2| = 2^x + x^2 \le 2^x + 2^x$$

I have no idea how to get this to become just $$2^x$$ or if I am even doing this right. Any guidance is appreciated.

• Do you know whether $2 f(x)$ is $O(f(x))$? What does the big "$O$" mean? Oct 22, 2019 at 15:37
• You don't have to make it into $2^x$. Note that the definition of $O(2^x)$ gives you another piece to work with. Oct 22, 2019 at 15:38
• @Arthur can you elaborate? Oct 22, 2019 at 15:40
• @Zevias Read the definition of $O$ one more time. You will see that $f(x)\in O(2^x)$ doesn't mean $f(x)\leq 2^x$. It means something a little different. And that little extra is exactly what you need to finish the proof here. Also, $2^x+x^2\leq |2^x|+|x^2|\leq 2^x+x^2$ looks a bit unnecessary. Oct 22, 2019 at 16:13
• @Arthur I still am not sure what to do. I reduced it further to 2 2^x so that means f(x) is O(2^x) with witness C = 2 and witness K = (-infinite, +infinity)? Oct 22, 2019 at 16:50

Following the definition, can you find constants $$M$$ and $$C$$ such that for all $$x>M$$,

$$2^x+x^2< C\,2^x\text{ ?}$$

In other words,

$$1+x^22^{-x}< C.$$

By a quick study of the function $$1+x^2e^{-x}$$, we see that it is decreasing for $$M\ge3$$. As the value at $$3$$ is $$1+\dfrac98$$, we have for instance

$$x>3\implies 2^x+x^2\ge3\cdot2^x.$$

We illustrate this on the plot below, where the vertical axis is logarithmic: