For the following function, find the intervals on which the function is increasing or decreasing. \( f(x)=\sqrt{x^{2}+7} \)

To find the intervals on which the function \( f(x) = \sqrt{x^2 + 7} \) is increasing or decreasing, we first need to calculate its derivative. The derivative \( f'(x) \) can be found using the chain rule: \[ f'(x) = \frac{1}{2\sqrt{x^2 + 7}} \cdot (2x) = \frac{x}{\sqrt{x^2 + 7}}. \] Now, we analyze the sign of \( f'(x) \). The function is increasing where \( f'(x) > 0 \) and decreasing where \( f'(x) < 0 \): - \( f'(x) > 0 \): This occurs when \( x > 0 \). - \( f'(x) < 0 \): This occurs when \( x < 0 \). Thus, the function \( f(x) \) is decreasing on the interval \( (-\infty, 0) \) and increasing on the interval \( (0, \infty) \). The critical point is at \( x = 0\), where the function transitions from decreasing to increasing, forming a local minimum.