Is $f(x) = \frac{3}{\ln⁡(x+2)}$ an increasing or decreasing function? Is $f(x) = \frac{3}{\ln⁡(x+2)}$ an increasing or decreasing function? And is it an invertible function?
I'm aware an increasing function is when $f'(x)>0$ but i wasn't sure of the answer with the function above?
 A: Taking the derivative of $f(x)$, we have : 
$$f'(x) = \bigg(  \frac{3}{\ln⁡(x+2)} \bigg)' =-\frac{3}{(x+2)\ln^2(x+2)}.$$
Keep in mind that, for your function to be well defined, you'll need
$$x+2 >0 \Leftrightarrow x > -2,$$
and
$$x+2 \neq 1 \Leftrightarrow x \neq -1.$$
But $\forall x>-2$, $f'(x) <0$, which means that your function is strictly decreasing.
Since your function is strictly decreasing, it's also $``1-1"$, which means there exists its inverse function $f^{-1}(x)$.
A: On what domain do you consider the function? I suppose $x>-1$ such that $\ln(x+2)>0$.
Compute $f'(x)=-\frac{3}{(x+2)\ln^2(x+2)}<0$  for all $x>-1$.
Hence $f$ is strictly decreasing, hence injective.
To talk about bijectivity you need a proper definition of $f$ with domain and range. Since $f$ is injective, it is bijective onto its image $f((-1,\infty))=(0,\infty)$.
A: The function $x\mapsto x+2$ is increasing, so $x\mapsto \ln(x+2)$ is increasing and thus $x\mapsto {3\over \ln(x+2)}$ is decreasing on $(-1,\infty)$ and on $(-2,-1)$.
A: yes it is increasing and invertible, to be sure the best way is to make a study for the function with a graph
for the inverse, let $$y=f(x) = \frac{3}{\ln⁡(x+2)}\iff$$  $$\ln⁡(x+2)= \frac{3}{y}\iff $$ $$x+2= e^\frac{3}{y} \iff$$
$$x= e^\frac{3}{y}-2 \iff$$
thus
$$f^{-1}(x)= e^\frac{3}{x}-2$$
A: $$f'(x) = -\frac{3}{(2 + x) \ln^2(2 + x)}$$ which looks like this:

Since $f'(x) < 0$ for all $x$, it is strictly decreasing, which means it is one-to-one (bijective) and therefore invertible.
