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

To determine where the function \( f(x) = \sqrt{x^2 + 4} \) is increasing or decreasing, we first need to calculate its derivative. The derivative \( f'(x) \) is given by: \[ f'(x) = \frac{d}{dx} \left(\sqrt{x^2 + 4}\right) = \frac{x}{\sqrt{x^2 + 4}} \] Next, we analyze the sign of \( f'(x) \). The numerator \( x \) is positive when \( x > 0 \) and negative when \( x < 0 \). The denominator \( \sqrt{x^2 + 4} \) is always positive since the expression inside the square root is always greater than or equal to 4. Thus, we find that: - For \( x < 0 \), \( f'(x) < 0 \) (the function is decreasing). - For \( x > 0 \), \( f'(x) > 0 \) (the function is increasing). At \( x = 0 \), \( f'(0) = \frac{0}{\sqrt{0 + 4}} = 0 \), which indicates a change in monotonicity but does not contribute to interval classification. In summary, the function \( f(x) = \sqrt{x^2 + 4} \) is decreasing on the interval \( (-\infty, 0) \) and increasing on the interval \( (0, \infty) \).