find the critical points of f'(x) = 3 - sqrt(x)

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 \(x\).

Find the critical points of f'(x) = 3 - sqrt(x)

Learn how to find the critical points of the derivative function f'(x) = 3 - sqrt(x). This detailed solution walks through setting the derivative to zero and analyzing where the function is undefined. Suitable for students in Grades 10-12.

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