Existence of the max I have to prove that I there exists the max of this function, and I have to find it.
$$\max_{x\in\mathbb{R}\backslash \{0\}}\{\ln\vert x \vert - x^2 + 4\} $$
So I thought I could calculate the derivative:
$$\dfrac{1}{\vert x \vert} \text{sgn(x)} - 2x$$
The zeroes are $x = \pm\sqrt{1/2}$ but the function in these values returns the same value. How to deal with this?
To prove there is a max, I observe that
$$\lim_{x\to \pm \infty} \ln\vert x \vert -x^2+4 = -\infty$$
Can I from here conclude that the function because of this has max?
Also, how can I prove that $\ln\vert x \vert - x^2 < -\frac{3}{4}$ for all $x$ in $\mathbb{R}-\{0\}$?
Thank you!
 A: $f(-x)=f(x)$ so if $f$ attains it maximum value at $x_0,$ it will also attain it at $-x_0.$
To conclude there is a max, you must also prove that $\lim_0f=-\infty.$
Then, the max is as you found : at $\pm\frac1{\sqrt2}.$
Therefore, $\ln\vert x \vert - x^2 < -\frac{3}{4}$ for all $x\in\mathbb{R}\setminus\{0\}$ is equivalent to $f\left(\frac1{\sqrt2}\right)<-\frac{3}{4}+4$ i.e. (after simplification) $\ln2>\frac12.$
A: As Anne mentioned, since the function is symmetric about $x=0$, you only need to analyze one side; e.g. $x \in [0,\infty)$. Then you can simply the $|x|$ to $x$.
One way to argue that a maximum exists is the following

*

*Explain why the function is continuous throughout the domain of interest. Since the function is not finite at 0, consider $(0,\infty)$ as the domain.

*Show that the limit of the function at the ends of the domain is either finite, or tends to $-\infty$.

*The above two properties should be sufficient to guarantee there exists a maximum. But for a homework/test, you would probably have to explain why they guarantee a maximum.

Note that solutions of derivative being 0 are not guaranteed to be maxima. To say they are maximum, you also need to show that the second derivative evaluated at that point is negative.
