Can someone please help with this proof. Note that $x \in \mathbb{R}, x > 1$
Here was my attempt...
Since $11/2 > 5$, we know that $x^5 \leq x^{11/2}$ (since $x > 1$). Also $9 \leq 9x^{11/2}$. So then $38x^5 + 9 \leq 38x^{11/2} + 9x^{11/2}$. And since $\frac{1}{2} < \frac{11}{2}$, we know that $x^{1/2} < x^{11/2}$ so $\sqrt{x} < x^{11/2}$. Thus, $\sqrt{x}(38x^5 + 9) \leq x^{11/2}(38x^{11/2} + 9x^{11/2})$.
However after simplification, I no longer have an $x^{11/2}$ on the right side, which I need to prove that the left side is in $\mathcal{O}(x^{11/2})$.
What am I doing wrong? Any help would be greatly appreciated. Thank you!
The most important observation is that $\sqrt{x} = x^{1/2}$, and then multiplying this into the parenthesis as the first step. Hence $$\sqrt{x}(38x^5 + 9) = 38x^{5+1/2} + 9x^{1/2} = 38x^{11/2} + 9x^{1/2}\leq 38x^{11/2} + 9x^{11/2} = 47x^{11/2}. $$