To find the derivative of the function $f(x) = 3x + 4$ at $a = 4$, we will use the definition of the derivative:

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

Step 1: Calculate $f(a)$

Given that $f(x) = 3x + 4$, for $a = 4$:

$f(4) = 3(4) + 4 = 12 + 4 = 16$

Step 2: Calculate $f(a + h)$

Now, we calculate $f(a + h)$, where $a = 4$:

$f(4 + h) = 3(4 + h) + 4 = 3(4) + 3h + 4 = 12 + 3h + 4 = 16 + 3h$

Step 3: Apply the limit definition of the derivative

Now, plug $f(4 + h)$ and $f(4)$ into the definition of the derivative:

$f'(4) = \lim_{h \to 0} \frac{f(4 + h) - f(4)}{h} = \lim_{h \to 0} \frac{(16 + 3h) - 16}{h}$

Simplify the expression:

$f'(4) = \lim_{h \to 0} \frac{3h}{h} = \lim_{h \to 0} 3 = 3$

Conclusion:

The derivative of $f(x) = 3x + 4$ at $a = 4$ is:

$f'(4) = 3$

Would you like further details or have any questions?

Here are 5 related questions to deepen understanding:

1. How would the derivative change if $f(x) = 5x + 2$?
2. What is the geometric interpretation of the derivative at a point?
3. How does the derivative of a constant function behave?
4. Can you find the derivative of a quadratic function, like $f(x) = x^2$, using the definition of the derivative?
5. What does it mean if the derivative is zero at a point?

Tip: The derivative represents the slope of the tangent line to the graph of the function at a given point.