We are given the expression for the derivative:

$\lim_{h \to 0} \frac{7 - 2(2 + h) - 3}{h}$

Step 1: Simplify the expression inside the limit

First, distribute the $2$ inside the parentheses:

$7 - 2(2 + h) - 3 = 7 - (4 + 2h) - 3 = 7 - 4 - 2h - 3 = 0 - 2h = -2h$

So, the expression becomes:

$\lim_{h \to 0} \frac{-2h}{h}$

Step 2: Simplify the fraction

$\frac{-2h}{h} = -2$

Thus, the expression simplifies to:

$\lim_{h \to 0} (-2) = -2$

Step 3: Interpretation as a derivative

The given expression represents the derivative of some function $f(x)$ at a point $x = c$. Based on the limit form, this is the difference quotient:

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

By comparing this to the given expression, we can identify:

• $f(c + h) = 7 - 2(2 + h)$
• $f(c) = 7 - 2(2) = 3$

Thus, the function is:

$f(x) = 7 - 2x$

The point where we are evaluating the derivative is $c = 2$, since the expression inside the limit involves $2 + h$.

Conclusion

• The function is $f(x) = 7 - 2x$
• The value of $c$ is $2$

The derivative $f'(x) = -2$, and indeed the limit confirms that the derivative at $x = 2$ is $-2$.

---

Would you like further details on any part of the explanation?

Here are 5 related questions:

1. How do you differentiate other linear functions like $f(x) = 3x + 4$?
2. Can the derivative of $f(x) = 7 - 2x$ change if the function was nonlinear?
3. What does the value of the derivative tell us about the graph of $f(x)$?
4. How would you approach the derivative of a quadratic function?
5. How do limits relate to the concept of continuity?

Tip: The derivative of a constant linear function $f(x) = ax + b$ is always equal to the coefficient of $x$, which is $a$.