The function $g(x)=3x+\ln2x$ for $x>0$. Find $g'(x)$ and prove that $g(x)$ has an inverse function. 
The function $g(x)=3x+\ln2x$ for $x>0$. Find $g'(x)$ and prove that $g(x)$ has an inverse function.

So $g'(x)=3+\frac{1}{x} $ for $x>0$ But I have no idea how I'm supposed to prove that this means $g(x)$ has an inverse function. Any help will be appreciated 
 A: What is required for a function to be invertible? It must be one-to-one and onto. Your function is $g:\mathbb{R}^+\rightarrow\mathbb{R}$. Establishing one-to-one-ness should be doable via the derivative you calculated, and establishing onto-ness should follow from continuity, the intermediate value theorem, and the appropriate limits to $0^+$ and $\infty$.
A: Hint: It follows from what you did that $x>0\implies g'(x)>0$.
A: Since you have the condition $x>0$ and $g'(x) > 0$ for positive $x$, then you know $g$ is strictly increasing. This means that $g$ is injective on $(0,\infty)$ and has an inverse on that interval.
A: Generally speaking, there are two `standard' ways to show the existence of something. The first is to construct it, and the second is to show that if we suppose that such a thing does not exist, we get a contradiction. The intuitionist in me would always recommend the former before falling back on the latter. In this case you need to show that $g'(x)=3+\frac{1}{x}$. To find the inverse of a function $f$, I was taught to substitute to let $y=f(x)$, swap the roles of $x$ and $y$, then rearrange for $y$. In this case, this amounts to setting $x=g'(y)=3+\frac{1}{y}$ and rearranging for $y$. Good luck!  
EDIT:
It's been pointed out that you've been asked to find the inverse of $g$ and not of $g'$. In that case, we recall the following lemma:

Let $f:I\to \mathbb{R}$ be a strictly monotonic, continuous function on an interval $I$. Then $f$ has an inverse.

In light of this lemma, it suffices to show that $g$ is monotonic (since $\mathbb{R}^+=(0,\infty)$ is an interval). To do so we require that $g'(x)>0$ (or $g'(x)<0$) $\forall x\in \mathbb{R}^+$. The result follows quickly from you calculation.
