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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:

enter image description here

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).

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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 $x = 0$?

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  • $\begingroup$ But the problem is that if I perform a similar procedure on the $x < 0$ half like you said, I do get a local extremum point at $-\dfrac{1} {4}$ and I don't know what's wrong. I shouldn't get a minimum or maximum at that value. $\endgroup$ – user1502 Dec 16 '19 at 21:14
  • $\begingroup$ @user1502 : $x = -1/4$ is not a solution of $-2\sqrt{-x} - 1 = 0$. $\endgroup$ – Eric Towers Dec 16 '19 at 21:27
  • $\begingroup$ Just one more question: Isn't there also something to consider for a value like $x = \dfrac{3+\sqrt{5}}{2}$ since for that value of $x$ the derivative of $f(x)$ is again not defined (denomiantor = 0, because of the term $1+\sqrt{x}-x$). Wouldn't the function have an extremum at this point also? Just like in the case where $x=0$, the case you told me to look at in the last sentence of your answer. $\endgroup$ – user1502 Dec 17 '19 at 23:54
  • $\begingroup$ @user1502 : No extremum. No change of sign in the derivative as $x$ crosses that value. In fact, the function isn't even defined to the right of that point. $\endgroup$ – Eric Towers Dec 18 '19 at 0:02

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