How do you determine the intervals for which the function is increasing or decreasing given$f\left( x \right) = \left| {{x^2} - 4} \right|$?

Hint: First we have to split the function into two functions depending on the intervals. Then, we have to differentiate both the terms with respect to $x$, that is, we have to find $f'\left( x \right)$. If $f'\left( x \right) > 0$, then the function is increasing in that interval and if $f'\left( x \right) < 0$, then the function is decreasing in that interval. We must keep in mind that, we have to split the given modulus function because the modulus function is not differentiable, so we have to split the function into intervals and then differentiate.

Complete step by step solution:
Given, function is, $f\left( x \right) = \left| {{x^2} - 4} \right|$.
Now, we can write, ${x^2} - 4 = \left( {x - 2} \right)\left( {x + 2} \right)$.
Therefore, $f\left( x \right) = \left| {\left( {x - 2} \right)\left( {x + 2} \right)} \right|$
Now, we know, the modulus function is always positive.
Therefore, $\left( {x - 2} \right)\left( {x + 2} \right) \geqslant 0$
Now, using the method of intervals on the real line, we get,

So, on the real line, for positive values, the value of $x$ have to be within the intervals which are indicated by + sign in the real line, that is,
$x \in ( - \infty , - 2] \cup [2,\infty )$
And, for negative values, the value of $x$ have to be within the intervals which are indicated by - sign in the real line, that is,
$x \in ( - 2,2)$
Therefore, we can split the function, in the following way,
In the interval $x \in ( - \infty , - 2] \cup [2,\infty )$, the function is,
$f\left( x \right) = {x^2} - 4$
And, in the interval $x \in ( - 2,2)$, the function is,
$f\left( x \right) = - ({x^2} - 4)$
Now, to find the intervals in which the function is increasing and decreasing let us differentiate the functions with respect to $x$.
Therefore, in the interval, $x \in ( - \infty , - 2] \cup [2,\infty )$
$f'\left( x \right) = \dfrac{d}{{dx}}\left( {{x^2} - 4} \right)$
$ \Rightarrow f'\left( x \right) = 2x$
Now, for the interval, $( - \infty , - 2]$,
$f'\left( x \right) < 0$
Hence, the function is decreasing in $( - \infty , - 2]$.
And, for the interval, $[2,\infty )$,
$f'\left( x \right) > 0$
Hence, the function is increasing in $[2,\infty )$.
Now, for the interval $x \in ( - 2,2)$,
$f'\left( x \right) = \dfrac{d}{{dx}}\left( { - ({x^2} - 4)} \right)$
$ \Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left( { - {x^2} + 4)} \right)$
$ \Rightarrow f'\left( x \right) = - 2x$
Now, in the interval, $( - 2,0)$,
$f'\left( x \right) > 0$
Hence, the function is increasing in $( - 2,0)$.
And, in the interval, $(0,2)$,
$f'\left( x \right) < 0$
Hence, the function is decreasing in $(0,2)$.
Therefore, we can say that the function increases in the interval $( - 2,0) \cup [2,\infty )$. And, decreasing in the interval $( - \infty , - 2] \cup (0,2)$.

Note: The problem could also be solved by checking the various intervals of the functions after splitting the modulus function and then checking with various values of the defined intervals $( - \infty , - 2],( - 2,0),(0,2),[2,\infty )$ and see the trend of the values of the function un these intervals and hence find the solution to the question.