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We are tasked with finding and simplifying the **difference quotient** for the function \( f(x) = 7x^2 \). The difference quotient is given by:

\[
\frac{f(x+h) - f(x)}{h}, \quad h \neq 0
\]

### Step-by-step solution:

1. **Substitute \( f(x) = 7x^2 \) into the formula:**

   \[
   f(x+h) = 7(x+h)^2
   \]

2. **Expand \( f(x+h) \):**

   \[
   f(x+h) = 7(x^2 + 2xh + h^2) = 7x^2 + 14xh + 7h^2
   \]

3. **Now compute \( f(x+h) - f(x) \):**

   \[
   f(x+h) - f(x) = (7x^2 + 14xh + 7h^2) - 7x^2 = 14xh + 7h^2
   \]

4. **Divide by \( h \):**

   \[
   \frac{f(x+h) - f(x)}{h} = \frac{14xh + 7h^2}{h} = 14x + 7h
   \]

Thus, the simplified difference quotient is:

\[
\frac{f(x+h) - f(x)}{h} = 14x + 7h
\]

This expression is simplified, but as \( h \to 0 \), it will approach \( 14x \), which is the derivative of the function \( f(x) \).

Would you like further details on any step?

### Related questions:
1. What is the derivative of \( f(x) = 7x^2 \) using limits?
2. How is the difference quotient related to the concept of a derivative?
3. How does the difference quotient change for different powers of \( x \) in functions?
4. What happens to the difference quotient when \( h \to 0 \)?
5. Can the difference quotient be used to approximate instantaneous rates of change?

### Tip:
The difference quotient is a foundational concept in calculus, as it directly leads to the formal definition of a derivative.

Find and simplify the difference quotient f(x+h) - f(x) / h, h ≠ 0 for the given function. f(x) = 7x^2.

Learn how to find and simplify the difference quotient for the quadratic function f(x) = 7x^2. This solution involves expanding binomials and simplifying algebraic expressions, a key concept in calculus. Suitable for students in Grades 11-12.

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