Given
f ( x )

To find $f'(1)$ using the definition of a derivative, we use the limit definition of the derivative:

$f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$

In this case, $f(x) = 3x - 2$, and we are asked to find $f'(1)$. So, we set $a = 1$ and proceed step by step.

Step 1: Substitute $f(x) = 3x - 2$ and $a = 1$ into the limit definition:

$f'(1) = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h}$

Step 2: Compute $f(1+h)$ and $f(1)$

• $f(1+h) = 3(1+h) - 2 = 3 + 3h - 2 = 1 + 3h$
• $f(1) = 3(1) - 2 = 1$

Step 3: Substitute these into the limit expression:

$f'(1) = \lim_{h \to 0} \frac{(1 + 3h) - 1}{h}$

Step 4: Simplify the expression:

$f'(1) = \lim_{h \to 0} \frac{1 + 3h - 1}{h} = \lim_{h \to 0} \frac{3h}{h}$

Step 5: Cancel $h$ and compute the limit:

$f'(1) = \lim_{h \to 0} 3 = 3$

Final Answer:

$f'(1) = 3$

Would you like more details, or do you have any questions?

Here are 5 related questions:

1. How would you find $f'(2)$ using the same method?
2. What is the geometric meaning of the derivative at $x = 1$?
3. How does the derivative of $f(x) = 3x - 2$ compare to the slope of the line?
4. Can you apply the definition of the derivative to a nonlinear function like $f(x) = x^2$?
5. What happens to the derivative if the function is constant?

Tip: The derivative of a linear function like $f(x) = 3x - 2$ is always the slope of the line, which remains constant for all $x$.