The question asks for the maximum value of the function $\sqrt{x}$ on the interval $I = [1, 4]$.

To solve this, let's analyze the function:

• The function $f(x) = \sqrt{x}$ is increasing because as $x$ increases, $\sqrt{x}$ also increases.
• On the interval $[1, 4]$, the minimum value of $f(x)$ occurs at $x = 1$, which gives $\sqrt{1} = 1$.
• The maximum value of $f(x)$ occurs at $x = 4$, which gives $\sqrt{4} = 2$.

Therefore, the maximum value of $\sqrt{x}$ on the interval $[1, 4]$ is 2, which corresponds to option e.

Would you like further details or clarifications?

Here are 5 related questions to deepen understanding:

1. What is the domain of the function $\sqrt{x}$?
2. How does the function $\sqrt{x}$ behave as $x$ approaches infinity?
3. What would the maximum value be if the interval was $[0, 4]$?
4. How would the answer change if we were looking for the minimum value instead?
5. Can you explain how the derivative of $\sqrt{x}$ confirms that it is an increasing function?

Tip: The square root function is defined only for non-negative numbers, and it increases monotonically for $x \geq 0$.