how many real numbers does. the domain of the fanction \( f \) not included? \[ f_{(x)}=\frac{x+2}{x^{2}-5|x|+4} \]

To find the domain of the function \( f(x) = \frac{x+2}{x^{2}-5|x|+4} \), we need to consider the values of \( x \) for which the denominator is not equal to zero. The denominator of the function is \( x^{2}-5|x|+4 \). To find the values of \( x \) for which the denominator is not equal to zero, we need to solve the equation \( x^{2}-5|x|+4 = 0 \). Let's solve the equation \( x^{2}-5|x|+4 = 0 \) to find the values of \( x \) that make the denominator zero. Solve the equation by following steps: - step0: Solve for \(x\): \(x^{2}-5\left|x\right|+4=0\) - step1: Separate into possible cases: \(\begin{align}&x^{2}-5x+4=0,x\geq 0\\&x^{2}-5\left(-x\right)+4=0,x<0\end{align}\) - step2: Solve the equation: \(\begin{align}&\begin{align}&x=4\\&x=1\end{align},x\geq 0\\&\begin{align}&x=-1\\&x=-4\end{align},x<0\end{align}\) - step3: Find the intersection: \(\begin{align}&x=4\\&x=1\\&x=-1\\&x=-4\end{align}\) - step4: Rewrite: \(x_{1}=-4,x_{2}=-1,x_{3}=1,x_{4}=4\) The solutions to the equation \( x^{2}-5|x|+4 = 0 \) are \( x = -4, -1, 1, 4 \). These values of \( x \) make the denominator zero, and therefore, they are not included in the domain of the function \( f(x) = \frac{x+2}{x^{2}-5|x|+4} \). Therefore, the domain of the function \( f(x) \) does not include the real numbers \( x = -4, -1, 1, 4 \).