The question asks to use the four-step process to find the derivative $f'(x)$ of the function $f(x) = x^2 + 3x - 2$ and then to find $f'(1)$, $f'(2)$, and $f'(3)$.

The four-step process for finding the derivative is:

1. Step 1: Find $f(x + h)$, where $f(x) = x^2 + 3x - 2$.

2. Step 2: Compute the difference $f(x + h) - f(x)$.

3. Step 3: Divide the difference by $h$, i.e., find $\frac{f(x + h) - f(x)}{h}$.

4. Step 4: Take the limit as $h \to 0$ of the expression from Step 3 to get $f'(x)$.

Let's apply this step by step.

Step 1: Find $f(x + h)$

$f(x + h) = (x + h)^2 + 3(x + h) - 2$ Expanding the terms: $f(x + h) = (x^2 + 2xh + h^2) + (3x + 3h) - 2$ $f(x + h) = x^2 + 2xh + h^2 + 3x + 3h - 2$

Step 2: Compute $f(x + h) - f(x)$

Now, subtract $f(x) = x^2 + 3x - 2$ from $f(x + h)$: $f(x + h) - f(x) = (x^2 + 2xh + h^2 + 3x + 3h - 2) - (x^2 + 3x - 2)$ Simplifying: $f(x + h) - f(x) = 2xh + h^2 + 3h$

Step 3: Divide by $h$

Now, divide the expression by $h$: $\frac{f(x + h) - f(x)}{h} = \frac{2xh + h^2 + 3h}{h}$ Simplifying: $\frac{f(x + h) - f(x)}{h} = 2x + h + 3$

Step 4: Take the limit as $h \to 0$

Now, take the limit as $h \to 0$: $f'(x) = \lim_{h \to 0} (2x + h + 3) = 2x + 3$

Thus, the derivative $f'(x) = 2x + 3$.

Now, find $f'(1)$, $f'(2)$, and $f'(3)$:

• $f'(1) = 2(1) + 3 = 2 + 3 = 5$
• $f'(2) = 2(2) + 3 = 4 + 3 = 7$
• $f'(3) = 2(3) + 3 = 6 + 3 = 9$

Final Answers:

• $f'(1) = 5$
• $f'(2) = 7$
• $f'(3) = 9$

Would you like more details on any of the steps or have any further questions?

Here are five related questions to explore further:

1. How would the four-step process change for a function with a higher degree, like $f(x) = x^3 + 2x^2 - x$?
2. How is the four-step process related to the definition of a derivative in calculus?
3. What would the second derivative $f''(x)$ be for this function?
4. How would the derivative change if the constant term was removed from the function?
5. What happens if you evaluate the derivative at negative values of $x$, like $f'(-1)$?

Tip: The four-step process is a formal way to derive the derivative using the definition of limits, but you can often take shortcuts using known derivative rules.