To find the critical points of the function $f'(x) = 3 - \sqrt{x}$, we need to set the derivative equal to zero and solve for $x$, as critical points occur where the derivative is zero or undefined.

Step 1: Set the derivative equal to zero

$3 - \sqrt{x} = 0$ Solve for $\sqrt{x}$: $\sqrt{x} = 3$ Square both sides to eliminate the square root: $x = 9$

Step 2: Check where the derivative is undefined

The derivative $f'(x) = 3 - \sqrt{x}$ is undefined when $x < 0$ because $\sqrt{x}$ is undefined for negative numbers in the real number system. Therefore, $x \geq 0$ is the domain of the derivative.

Conclusion:

The critical point is at $x = 9$. This is where the derivative is zero.

Would you like a detailed explanation of how to analyze whether this critical point is a minimum, maximum, or neither?

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Here are five related questions:

1. How can we determine if $x = 9$ is a local minimum or maximum?
2. What is the original function $f(x)$ if $f'(x) = 3 - \sqrt{x}$?
3. How does the second derivative test help in determining the nature of critical points?
4. Can there be any inflection points in the function $f(x)$?
5. How would the behavior of the function change for values $x > 9$?

Tip: Always check the domain of a function when dealing with square roots or other operations that restrict possible values for $x$.