We need to find the expression for $\frac{f(3+h) - f(3)}{h}$ given the function $f(x) = x^2 + 5$.

Step 1: Calculate $f(3+h)$

Substitute $x = 3+h$ into the function $f(x) = x^2 + 5$: $f(3+h) = (3+h)^2 + 5$ Expand $(3+h)^2$: $(3+h)^2 = 9 + 6h + h^2$ So, $f(3+h) = 9 + 6h + h^2 + 5 = h^2 + 6h + 14$

Step 2: Calculate $f(3)$

Substitute $x = 3$ into $f(x) = x^2 + 5$: $f(3) = 3^2 + 5 = 9 + 5 = 14$

Step 3: Compute $\frac{f(3+h) - f(3)}{h}$

Now subtract $f(3)$ from $f(3+h)$: $f(3+h) - f(3) = (h^2 + 6h + 14) - 14 = h^2 + 6h$ Now, divide by $h$: $\frac{f(3+h) - f(3)}{h} = \frac{h^2 + 6h}{h}$ Simplify the expression: $\frac{f(3+h) - f(3)}{h} = h + 6$

Final Answer:

$\boxed{h + 6}$

Would you like to explore this further or have any questions?

Here are some related questions you might find interesting:

1. How would you find the derivative of $f(x) = x^2 + 5$ using the limit definition?
2. What happens to the expression $\frac{f(3+h) - f(3)}{h}$ as $h$ approaches 0?
3. How do you calculate $\frac{f(x+h) - f(x)}{h}$ for a general function $f(x)$?
4. What is the value of $\frac{f(3+h) - f(3)}{h}$ when $h = 1$?
5. How would the solution change if $f(x)$ were $x^2 - 5$ instead of $x^2 + 5$?

Tip: Expressions like $\frac{f(x+h) - f(x)}{h}$ are fundamental in calculus, as they form the basis for understanding derivatives.