For the following function, find the intervals on which the function is increasing or decreasing. \( f(x)=\sqrt{x^{2}+4} \)

Analyze the function features with the analysis type as "intervals" for the function \( f(x)=\sqrt{x^{2}+4} \). Function by following steps: - step0: Find the increasing or decreasing interval: \(f\left(x\right)=\sqrt{x^{2}+4}\) - step1: Find the domain: \(f\left(x\right)=\sqrt{x^{2}+4},x \in \mathbb{R}\) - step2: Find the derivative: \(f^{\prime}\left(x\right)=\frac{x}{\sqrt{x^{2}+4}}\) - step3: Find the domain: \(f^{\prime}\left(x\right)=\frac{x}{\sqrt{x^{2}+4}},x \in \mathbb{R}\) - step4: Substitute \(f^{\prime}\left(x\right)=0:\) \(0=\frac{x}{\sqrt{x^{2}+4}}\) - step5: Swap the sides: \(\frac{x}{\sqrt{x^{2}+4}}=0\) - step6: Cross multiply: \(x=\sqrt{x^{2}+4}\times 0\) - step7: Simplify the equation: \(x=0\) - step8: Determine the intervals: \(\begin{align}&x\leq 0\\&x\geq 0\end{align}\) - step9: Choose the points: \(\begin{align}&x_{1}=-1\\&x_{2}=1\end{align}\) - step10: Find the values of the derivatives: \(\begin{align}&f^{\prime}\left(-1\right)\approx -0.447214\\&f^{\prime}\left(1\right)\approx 0.447214\end{align}\) - step11: Calculate: \(\begin{align}&x\leq 0\textrm{ is decreasing interval}\\&x\geq 0\textrm{ is increasing interval}\end{align}\) - step12: Evaluate: \(\begin{align}&\textrm{The increasing interval is}\textrm{ }x\geq 0\\&\textrm{The decreasing interval is}\textrm{ }x\leq 0\end{align}\) The function \( f(x)=\sqrt{x^{2}+4} \) is increasing on the interval \( x \geq 0 \) and decreasing on the interval \( x \leq 0 \).