To calculate the derivative $f'(x)$ of the function $f(x) = 3x^2 - 7$ using the limit definition of a derivative, we follow these steps.

Limit Definition of Derivative:

The derivative of a function $f(x)$ is given by:

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

Step-by-Step Solution:

1. Write down the function: $f(x) = 3x^2 - 7$

2. Substitute $f(x+h)$ into the limit definition: We first need to find $f(x+h)$.

   $f(x + h) = 3(x+h)^2 - 7$

   Expand $(x+h)^2$:

   $(x+h)^2 = x^2 + 2xh + h^2$

   Now substitute this into the expression for $f(x + h)$:

   $f(x + h) = 3(x^2 + 2xh + h^2) - 7 = 3x^2 + 6xh + 3h^2 - 7$

3. Set up the difference quotient:

   Now calculate the difference $f(x+h) - f(x)$:

   $f(x+h) - f(x) = \left( 3x^2 + 6xh + 3h^2 - 7 \right) - \left( 3x^2 - 7 \right)$

   Simplifying this:

   $f(x+h) - f(x) = 3x^2 + 6xh + 3h^2 - 7 - 3x^2 + 7 = 6xh + 3h^2$

4. Divide by $h$ to form the difference quotient:

   $\frac{f(x+h) - f(x)}{h} = \frac{6xh + 3h^2}{h}$

   Simplify by dividing each term by $h$:

   $\frac{f(x+h) - f(x)}{h} = 6x + 3h$

5. Take the limit as $h \to 0$:

   $f'(x) = \lim_{h \to 0} (6x + 3h)$

   As $h \to 0$, the term $3h$ vanishes, leaving:

   $f'(x) = 6x$

Final Answer:

The derivative of $f(x) = 3x^2 - 7$ is:

$f'(x) = 6x$

Would you like any further clarification, or details on any step?

Expanding Questions:

1. How would the process change if the function was $f(x) = 3x^2 - 7x + 2$?
2. Can you apply the limit definition of a derivative to a trigonometric function, such as $f(x) = \sin(x)$?
3. What is the geometric interpretation of the derivative you just calculated?
4. How would the second derivative of $f(x) = 3x^2 - 7$ be calculated using the same limit definition?
5. What does the derivative $f'(x) = 6x$ tell you about the slope of the tangent line to the curve $y = 3x^2 - 7$?

Tip:

Always expand $(x+h)^n$ carefully when using the limit definition to avoid algebraic mistakes.