To simplify the difference quotient for the function $f(x) = x^2 + 4$, we use the general form of the difference quotient:

$\frac{f(x+h) - f(x)}{h}.$

Steps:

1. Substitute $f(x)$ and $f(x+h)$:

   • $f(x) = x^2 + 4$,
   • $f(x+h) = (x+h)^2 + 4$.

   Expanding $f(x+h)$: $f(x+h) = (x+h)^2 + 4 = x^2 + 2xh + h^2 + 4.$

2. Set up the difference quotient: $\frac{f(x+h) - f(x)}{h} = \frac{\left( x^2 + 2xh + h^2 + 4 \right) - \left( x^2 + 4 \right)}{h}.$

3. Simplify the numerator:

   • Combine like terms and cancel out: $\left( x^2 + 2xh + h^2 + 4 \right) - \left( x^2 + 4 \right) = 2xh + h^2.$

   So the expression becomes: $\frac{2xh + h^2}{h}.$

4. Factor $h$ out of the numerator: $\frac{2xh + h^2}{h} = \frac{h(2x + h)}{h}.$

5. Cancel $h$ (as long as $h \neq 0$): $2x + h.$

Final Simplified Difference Quotient:

$2x + h.$

This is the simplified form of the difference quotient for $f(x) = x^2 + 4$.

Would you like further details on any step or need help applying this to a related problem?

Here are 5 follow-up questions you might find helpful:

1. How do you interpret the difference quotient geometrically?
2. What happens to the difference quotient as $h \to 0$?
3. How does the simplified difference quotient relate to the derivative of $f(x)$?
4. Can you simplify the difference quotient for other functions, such as $f(x) = x^3$?
5. How does this process change for linear functions like $f(x) = 2x + 1$?

Tip: The difference quotient is a foundational concept for derivatives, so simplifying it helps when working on calculus problems.