To determine the interval where $f(x) = \int_x^3 (3-t)^{1/2} \, dt$ is increasing, we analyze its derivative $f'(x)$. By the Fundamental Theorem of Calculus, the derivative of $f(x)$ with respect to $x$ is given by:

$f'(x) = -(3-x)^{1/2}$

The negative sign comes from the fact that the lower limit of the integral is $x$, not a constant. To find the intervals where $f(x)$ is increasing, we determine where $f'(x) > 0$.

Analyze $f'(x)$:

$f'(x) = -(3-x)^{1/2}$

1. The square root $(3-x)^{1/2}$ is defined only when $3 - x \geq 0$, or equivalently, $x \leq 3$. Outside this domain, $f(x)$ is not defined.

2. Since $(3-x)^{1/2} \geq 0$ for $x \leq 3$, the negative sign in $f'(x)$ implies that $f'(x) \leq 0$ for all $x \leq 3$.

Conclusion:

The function $f(x)$ is never increasing on its domain because $f'(x) \leq 0$ for all $x \leq 3$. It is either constant or strictly decreasing.

Would you like a more detailed breakdown or visualization of this result? Here are five related questions for further exploration:

1. How can we verify the decreasing nature of $f(x)$ explicitly from the integral definition?
2. What is the behavior of $f(x)$ as $x \to 3$?
3. Can you find the exact value of $f(x)$ at specific points?
4. How does changing the limits of integration affect whether $f(x)$ is increasing or decreasing?
5. What happens if the square root function is replaced by another power, such as $(3-t)^n$?

Tip: When analyzing integral functions, always start by applying the Fundamental Theorem of Calculus to determine the derivative, as it provides insights into monotonicity.