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.

Problem 1: $f(x) = 9$

Since $f(x)$ is a constant function:

• $f(x+h) = 9$
• $f(x+h) - f(x) = 0$

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

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 $h \to 0$, $f'(x) = -6$.

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 $h \to 0$: $f'(x) = 4x - 7$.

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