I am given the following function:
$$f: D \rightarrow \mathbb{R} \hspace{2cm} f(x) = \ln \bigg (1 + \sqrt{|x|} -x \bigg )$$
where $D$ is the maximum domain of the function. I have to find the number of local extrema points of this function. What I did was to first make the function look a little cleaner by getting rid of the absolute value:
$$ f(x) = \left\{ \begin{array}{ll} \ln(1 + \sqrt{x} - x) & \quad x \ge 0 \\ \ln(1 + \sqrt{-x}-x) & \quad x < 0 \end{array} \right. $$
I know that we have local extrema points where the derivative of the function is $0$ while the derivative approaching that point from right and left has opposite signs.
We can also have a local extremum in a point where the derivative does not exist (an example being the function $h(x) = |x|$). So naturally I found the derivative of the function:
$$ f'(x) = \left\{ \begin{array}{ll} \ \dfrac{1-2\sqrt{x}}{2\sqrt{x}(1+\sqrt{x}-x)} & \quad x > 0 \\ \\ \dfrac{1-2\sqrt{-x}}{2\sqrt{-x}(1+\sqrt{-x}-x)} & \quad x < 0 \end{array} \right. $$
But I got stuck here. I don't know how to continue. What bothers me is that I do not know the domain $D$ and I don't know how can I find it. If I try to find the values of $x$ for which
$$f'(x)=0$$
I get $x_0= \dfrac{1}{4}$ and $x_1 = -\dfrac{1}{4}$, but if I look at the graph of the function:
I can see that we don't have a local extremum point at $x_1 = -\dfrac{1}{4}$, even thought we do have a local extremum point at $x_0 = \dfrac{1}{4}$. By looking at the graph I also see that we have an extremum point at $x = 0$. I'm guessing that is because of the denominator of the derivative makes it such that the derivative is not defined at $x=0$, so we have an extremum point, but shouldn't we also take into consideration the values of $x$ for which the other term in the denominator of the derivative, $(1+\sqrt{\pm x} - x)$ is not defined? And how would we handle that? And how would I find that we have a local extremum point at $x=0$ analytically, without looking at the graph of the function? Also, isn't that point of the graph that is close to $y=-35$ also a local extremum point?
TL;DR: How can I find all the local extremum points of the given function without looking at the graph? The correct answer is $2$ (according to my textbook).
For a fraction to be zero, the numerator must be zero (with one caveat, discussed at the end of this paragraph). So you want to solve $$ 1 - 2 \sqrt{x} = 0 \text{,} $$ restricted to $x > 0$ and check each solution to see if it also makes the denominator zero. (If it does also make the denominator zero, take a limit approaching that potential solution to see what is really happening.)
This says $\sqrt{x} = 1/2$, so $x = 1/4$. Putting that in the denominator, $2 \sqrt{1/4}(1+\sqrt{1/4} - 1/4) = 5/4$, so we do not need to take a limit. Notice that for $0<x<1$, the denominator of $f'$ is positive, so we need only check the sign of the numerator to the left and right of $1/4$ in $(0,1)$. Let's check the signs of the numerators of $f'(1/16)$ and $f'(1/2)$ : $1 - 2/4$ is positive and $1 - 2/\sqrt{2}$ is negative, so we found a local maximum.
Perform a similar procedure on the $x < 0$ half and find no local extrema on that half.
But you have missed something. What is $f'(0)$ (Is it even defined?) and does $f'$ change signs while crossing $x = 0$?