how to find the maximum value of a function

Since we're dealing with the absolute value of a function, we can interpret the absolute value as giving us the distance from the function to the $x$ axis independently from the direction. In other words, we just want to find the point that's furthest away from the $x$ axis, independently if it's above or below the axis since the distance (absolute value) is all we care about maximizing. With this in mind, it becomes clear that the maximum of $\lvert f(x) \rvert$ has to be either a maximum or a minimum of $f(x)$.

From here it's a fairly straightforward (although lengthy) calculation to find the result. We see that given $f(x) = \sqrt{x^4-3 x^2-6 x+13}-\sqrt{x^4+5 x^2+4}$, the extrema points are found when the derivative equals $0$, hence \begin{align*} \frac{d}{dx} \left(\sqrt{x^4-3 x^2-6 x+13}-\sqrt{x^4+5 x^2+4}\right)& = \frac{4x^3-6x-6}{2\sqrt{x^4-3 x^2-6 x+13}} - \frac{4x^3 +10x}{2\sqrt{x^4+5 x^2+4}} \tag{1} \end{align*} And by equating equation $(1)$ to $0$ we get $$ \frac{4x^3-6x-6}{2\sqrt{x^4-3 x^2-6 x+13}} = \frac{4x^3 +10x}{2\sqrt{x^4+5x^2+4}} $$ $$ \implies \frac{\left(2x^3-3x-3\right)^2}{x^4-3 x^2-6 x+13} = \frac{\left(2x^3 +5x\right)^2}{x^4+5 x^2+4} $$ $$ \implies \left(x^4+5 x^2+4\right)\left(4 x^6 - 12 x^4 - 12 x^3 + 9 x^2 + 18 x + 9\right) = \left(x^4-3 x^2-6 x+13 \right)\left(4 x^6 + 20 x^4 + 25 x^2 \right) $$ $$ \implies \color{blue}{4 x^{10}} \mathbin{\color{brown}{-}} \color{brown}{12 x^{8}} - 12 x^{7} + 9 x^{6} + 18 x^5 + 9x^4 + \color{green}{20 x^8} \mathbin{\color{purple}{-}} \color{purple}{60 x^6} - 60 x^5 + 45 x^4 + 90 x^3 + 45 x^2 + 16 x^6 - 48 x^4 - 48 x^3 + 36 x^2 + 72 x + 36 = \color{blue}{4 x^{10}} + \color{green}{20 x^8} + 25 x^6 \mathbin{\color{brown}{-}} \color{brown}{12 x^{8}} \mathbin{\color{purple}{-}} \color{purple}{60 x^6} - 75 x^4 -24 x^7 - 120 x^5 - 150 x^3 + 52 x^6 + 260 x^4 + 325 x^2 $$ $$ \implies 12 x^7 - 52 x^6 + 78 x^5 - 179 x^4 + 192 x^3 - 244 x^2 + 72 x + 36 =0 $$ $$ \implies (3 x^2 - 4 x + 6) (4 x^5 - 12 x^4 + 2 x^3 - 33 x^2 + 16 x + 6)=0 $$

This last implication can be verified using long division. Since we can easily show (using the quadratic formula, for example) that the roots of $3 x^2 - 4 x + 6$ are complex, we can rule them out since we're only interested in real solutions. This simplifies our problem to finding the roots of $$ 4 x^5 - 12 x^4 + 2 x^3 - 33 x^2 + 16 x + 6 $$ Unfortunately, this last expression can't be factored any further, and since there isn't a quintic formula to solve for the roots the best we can do is find the roots using numerical methods. By doing so, you find that the real roots are given by \begin{align} x_1 &\approx -0.24569\\ x_2 & \approx 0.687122\\ x_3 & \approx 3.44225 \end{align} However, if we actually plug in $x_2$ into equation $(1)$ we see that we don't get $0$ (in fact it evaluates to approximately $-3.18$). This in fact is a "fake" root, since it is indeed a root of an equation that we obtained later in our procedure to find the roots, but it's not a cero to our derivative, which is what we're interested in. This leaves us with two extrema points of our function $f(x)$: $-0.24569$ and $3.44225$. By evaluating the function $f(x)$ at these points we see that $$ \sqrt{(-0.24569)^4-3 (-0.24569)^2-6 (-0.24569)+13}-\sqrt{(-0.24569)^4+5 (-0.24569)^2+4} \approx 1.706 $$ $$ \sqrt{(3.44225)^4-3 (3.44225)^2-6 (3.44225)+13}-\sqrt{(3.44225)^4+5 (3.44225)^2+4} \approx -4.1146 $$ And recalling that, because of the absolute value, we're just interested in the absolute value of these quantities, we see that $$ \lvert f(3.44225) \rvert = \lvert -4.1146 \rvert = 4.1146 \mathbin{\color{red}{>}} 1.706 = \lvert f(-0.24569) \rvert $$ And so, we can conclude that $x \approx 3.44225$ is the maximum value.

We can visually check our result by plotting the function $\left| \sqrt{x^4-3 x^2-6 x+13}-\sqrt{x^4+5 x^2+4}\right|$ as follows

enter image description here

In conclusion, the final result for the maximum can be stated as

The maximum value of $\left| \sqrt{x^4-3 x^2-6 x+13}-\sqrt{x^4+5 x^2+4}\right| $ is given by the largest positive root of $4 x^5 - 12 x^4 + 2 x^3 - 33 x^2 + 16 x + 6$ and has an approximate value of $3.44225$.