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!

It is not too difficult to visualize the graph of
$f(x)=e^{-3x}-3e^{-2x}+1$
Let's start by throwing away the $+1$, as this just moves the graph up by one unit. And let's replace $x$ by $-x$ which reflects the graph in the $y$ axis i.e. swaps right and left. So let's define
$g(x) = f(-x)-1 = e^{3x}-3e^{2x}$
The graph of $g(x)$ crosses the $y$ axis at $(0, -2)$. It also crosses the $x$ axis just once, when $e^{3x}=3e^{2x}$, so at $(\ln(3),0)$.
The $e^{3x}$ term dominates for large $x$, so the graph heads off into the upper right hand corner of the $x,y$ plane. And both terms in $g(x)$ are small for $x<<0$, so the graph approaches $y=0$ on the left hand side.
Finally $g'(x)=0$ when $3e^{3x}=6e^{2x}$, which has just one solution, when $x=\ln(2)$. So the graph has a single turning point at $(\ln(2), -4)$.
With this info, you can sketch the graph of $g(x)$ - it starts out just below the $x$ axis on the left hand side, heads downwards, crosses the $y$ axis at $(0,-2)$, carries on downwards to $(\ln(2), -4)$, then turns updwards, crosses the $x$ axis at $(\ln(3),0)$ and heads up and to the right in an exponential curve.
Reflect the sketch graph of $g(x)$ in the $y$ axis and add $1$ and you have a sketch graph of $f(x)$.
