Let's break down this problem and solve for each "blank."

Step 1: Function definition and difference quotient

You are given the function: $f(x) = 6x^2 + 9x$ The difference quotient formula for the instantaneous rate of change is: $\frac{f(x+h) - f(x)}{h}$ Substitute $f(x) = 6x^2 + 9x$ into this formula: $\frac{(6(x+h)^2 + 9(x+h)) - (6x^2 + 9x)}{h}$

Step 2: Expand $f(x+h)$

Now, let's expand $f(x+h) = 6(x+h)^2 + 9(x+h)$:

1. $6(x+h)^2 = 6(x^2 + 2xh + h^2) = 6x^2 + 12xh + 6h^2$
2. $9(x+h) = 9x + 9h$

So, $f(x+h) = 6x^2 + 12xh + 6h^2 + 9x + 9h$.

Step 3: Simplify the difference quotient

Now subtract $f(x) = 6x^2 + 9x$ from $f(x+h)$: $\frac{(6x^2 + 12xh + 6h^2 + 9x + 9h) - (6x^2 + 9x)}{h}$

Simplifying the numerator: $6x^2 + 12xh + 6h^2 + 9x + 9h - 6x^2 - 9x = 12xh + 6h^2 + 9h$

So, the expression becomes: $\frac{12xh + 6h^2 + 9h}{h}$

Step 4: Factor and cancel

Factor $h$ from the numerator: $\frac{h(12x + 6h + 9)}{h}$ Cancel the $h$ terms: $12x + 6h + 9$

Step 5: Take the limit as $h \to 0$

Now, take the limit as $h \to 0$: $\lim_{h \to 0} (12x + 6h + 9) = 12x + 9$

Final Answer for Blanks:

• First blank: $x$ (from $12x$)
• Second blank: $h$ (from $6h$)

Thus, the final expression is: $\lim_{h \to 0} (12x + 6h + 9) = 12x + 9$

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Let me know if you'd like further details or clarifications. Here are five related questions you might find useful:

1. What is the general formula for the derivative using limits?
2. How does the difference quotient relate to the derivative of a function?
3. What is the interpretation of the derivative in terms of rates of change?
4. What happens when you differentiate a constant function?
5. How does the derivative of a polynomial behave in general?

Tip: Always check your algebraic expansions carefully when calculating limits to avoid small errors!