The question is asking for the maximum value of $\sqrt{x}$ on the interval $I = (1, 4)$.

Solution:

The function given is $f(x) = \sqrt{x}$, which is defined and continuous for $x \geq 0$.

1. Domain of the interval: The interval is $I = (1, 4)$, which means the interval includes numbers from slightly greater than 1 up to 4, but does not include the endpoints 1 and 4.

2. Behavior of $f(x)$:

   • At $x = 1$, $\sqrt{1} = 1$.
   • At $x = 4$, $\sqrt{4} = 2$.

   The function $\sqrt{x}$ is an increasing function in the domain $x \geq 0$. Therefore, as $x$ approaches the upper bound of the interval (4), the function will reach its maximum value.

3. Maximum value: Since $x = 4$ is at the boundary of the interval and the function increases as $x$ increases, the maximum value of $\sqrt{x}$ on this interval is $\sqrt{4} = 2$.

Thus, the maximum value of $\sqrt{x}$ on the interval $I = (1, 4)$ is 2, which corresponds to option e.

Would you like more details on the reasoning behind this, or any specific part explained further?

Related Questions:

1. What is the derivative of $\sqrt{x}$, and how does it confirm the increasing nature of the function?
2. How would the answer change if the interval were closed $[1, 4]$?
3. Can $f(x) = \sqrt{x}$ have a minimum on the interval $(1, 4)$?
4. What is the difference between open and closed intervals in finding extrema?
5. How would the maximum change if the interval was $(0, 4)$?

Tip:

Always check if the interval is open or closed when finding maximum or minimum values, as it affects whether you can include the endpoints in your calculation.