Is the function $\ln(ax + b)$ increasing/decreasing, concave/convex? $h(x) = \ln(ax + b)$ 
NB. Examine your results acccording to values of $(a,b)$
I've differentiated twice in order to get the following:
$$
h''(x) = -\frac{a^2}{(ax+b)^2}
$$
I think this proves that $h(x)$ is concave for all value of $a$ and $b$ since $h''(x)\le0$. Is this correct?
I don't know how to prove whether it's increasing/decreasing or what the NB really means so any help with that would be great.
 A: we now that the domain of the function is:
$$ax+b\gt 0\Rightarrow x\gt\frac{-b}{a}\text{so the domain is:}(\frac{-b}{a},+\infty)$$
$$f'(x)=\frac{a}{ax+b}$$
in the domain of the function since we have $x\gt\frac{-b}{a}\Rightarrow ax+b>0 $ the sign of $f'(x)=\frac{a}{ax+b}$ will be dependent to the sign of $a$ so:
if $a\gt 0\Rightarrow f'(x)\gt 0$ and $f(x)$ will be increasing in its domain
if $a\lt 0\Rightarrow f'(x)\lt 0$ and $f(x)$ will be decreasing in its domain
Note the phrases in its domain in the above expression, we always study the behaviors of functions in their domain
$$f''(x)=\frac{-a^2}{(ax+b)^2}$$
as you said without we always have $f''(x)<0$ so without considering the sign of $a$, $f(x)$ will be a convex function
The value of $b$ doesnot influence the first and second derivation and so will not affect concavity, convexity, increase or decrease of the function 
here is the diagram for $f(x)=\ln(2x+1)$ as you see $a=2\gt 0$ and $f(x)$ is increasing and convex
 
here is the diagram for $f(x)=\ln(-2x+1)$ as you see $a=-2\lt 0$ and $f(x)$ is decreasing and convex

A: $$h^{ \prime  }\left( x \right) =\frac { a }{ ax+b } >0$$
$1$.if  $a>0$   $ax+b>0$  $\Rightarrow $ $ax>-b$ $\Rightarrow $ $x>-\frac { b }{ a } $ function is increasing
$2$.if   $a<0 $ $x<-\frac { b }{ a } $ function is decreasing
And about concavity you are right,find second derivative and check intervals
