Find intervals where $f$ is increasing/decreasing $f(x) = e^{-3x} -3e^{-2x} + 1$ I'm doing a problem where I'm asked to find the intervals where $f$ is decreasing or increasing of: 
$$f(x)=e^{-3x}-3e^{-2x}+1.$$
I've found that the derivative is $f'(x)=-3e^{-3x}+6e^{-2x}$ and that $f'(x)=0$ when $x=-\ln(2)$.
As far as I can see, $f$ is decreasing from $-\infty$ to $-\ln(2),$ and is increasing from $-\ln(2)$ to infinity. But $\lim_{x\to \infty}f(x) = 1$, which is what is confusing me now. 
Does this affect my answer? My thoughts are that  it is increasing as $x$ gets larger, just by a really tiny amount. And that $f'(x)$ approaches zero as $x$ gets bigger but it is never zero anywhere but $-\ln(2)$. Or am I just way off?
I'm also struggling with the follow up question: Find the largest interval $I$ that contains origo such that the function $g: I \to \mathbb{R}$ given by $g(x)=e^{-3x}-3e^{-2x}+1$ has an inverse function. 
If the interval has to contain $(0,0)$ then is the largest interval $\bigl(-\ln(2),\infty\bigr)$? How do I show this?
Thanks in advance!

Hint: Solve the inequality $$f'(x)\geq 0$$ this means $$-3e^{-3x}+6e^{-2x}\geq 0$$
This means $$e^x\geq \frac{1}{2}$$
