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To find the derivatives of each function, we’ll use the four-step process for differentiation:

1. **Rewrite \( f(x) \) if necessary (often not needed for simple polynomials)**
2. **Set up \( f(x+h) \) by replacing \( x \) with \( x+h \) in \( f(x) \)**
3. **Compute the difference quotient \( \frac{f(x+h) - f(x)}{h} \)**
4. **Take the limit as \( h \to 0 \) to find \( f'(x) \)**

Let's go through each function.

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### Problem 1: \( f(x) = 9 \)

Since \( f(x) \) is a constant function:
- \( f(x+h) = 9 \)
- \( f(x+h) - f(x) = 0 \)

Thus, \( f'(x) = 0 \).

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### Problem 2: \( f(x) = 4 - 6x \)

1. \( f(x+h) = 4 - 6(x+h) = 4 - 6x - 6h \)
2. \( f(x+h) - f(x) = (4 - 6x - 6h) - (4 - 6x) = -6h \)
3. Difference quotient: \( \frac{-6h}{h} = -6 \)
4. As \( h \to 0 \), \( f'(x) = -6 \).

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### Problem 3: \( f(x) = 2x^2 - 7x + 3 \)

1. \( f(x+h) = 2(x+h)^2 - 7(x+h) + 3 \)
   - Expanding: \( f(x+h) = 2(x^2 + 2xh + h^2) - 7x - 7h + 3 = 2x^2 + 4xh + 2h^2 - 7x - 7h + 3 \)
2. Difference quotient setup:
   - \( f(x+h) - f(x) = (2x^2 + 4xh + 2h^2 - 7x - 7h + 3) - (2x^2 - 7x + 3) = 4xh + 2h^2 - 7h \)
   - \( \frac{f(x+h) - f(x)}{h} = \frac{4xh + 2h^2 - 7h}{h} = 4x + 2h - 7 \)
3. Taking \( h \to 0 \): \( f'(x) = 4x - 7 \).

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### Problem 4: \( f(x) = \sqrt{x} + 2 \)

To differentiate \( f(x) = \sqrt{x} + 2 \), rewrite \( \sqrt{x} \) as \( x^{1/2} \).

1. \( f(x+h) = (x+h)^{1/2} + 2 \)
2. Difference quotient:
   - \( f(x+h) - f(x) = (x+h)^{1/2} + 2 - (\sqrt{x} + 2) = (x+h)^{1/2} - \sqrt{x} \)
   - This expression can be simplified using a rationalizing technique, giving \( \frac{1}{2\sqrt{x}} \).

The result

Use the four-step process to find f'(x) for the functions:
1) f(x) = 9
2) f(x) = 4 - 6x
3) f(x) = 2x^2 - 7x + 3
4) f(x) = sqrt(x) + 2

Learn to find the derivatives of functions using the four-step process. This includes examples with constant, linear, quadratic, and square root functions, covering key calculus concepts and formulas for differentiation. Suitable for high school seniors and college calculus students.

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