Stationary points of $ \ln(x + 1)$? I'm trying to find the stationary points of $f(x) =\ln(x + 1)$.
When I differentiate, I get $f'(x) =\dfrac{1}{ (x + 1)}.$
I then set that to zero and end up getting $1 = 0$?
I'm not sure what this means and I don't know what to do at this point. 
Any help? Thanks. 
 A: First, the domain of $f(x)= \ln(x+1)$ is $(-1, \infty)$. Furthermore, for all $x\in \mathbb R$, $\dfrac 1{x+1} \neq 0$.
That means that $f(x)$ has no minimum/maximum on the domain on which $\log(x+1)$ is defined.
To see this for yourself, graph $f(x) = \ln(x+1)$:

$f(x)$ is strictly increasing on the domain of definition.
A: Your differentiation is correct, and there is no solution for 
$$\frac{1}{1+x}=0,$$
for any $x$. What does this tell us about the graph of $f(x)=\ln(x+1)$? Well, since, mathematically, we cannot find any stationary points, it's worth looking at the graph of the function. Put the function into a graphics calculator, or type it into the input bar here. What we can see is that there seems to be no point at which the gradient of the function is zero, and therefore $\ln(x+1)$ does not have any stationary points.
As an interesting aside note, we know that 
$$\frac{d}{dx}e^x=e^x$$
and also that
$$e^x\neq 0 \ \ \ \text{for any} \ x$$
and hence $f(x)=e^x$ does not have any stationary points. Since $\ln(x)$ is the inverse function of $e^x$, we have an explanation as to why $ln(c+1)$ has no stationary points. 
