To find the difference quotient of the function $f(x) = \sqrt{17x}$, we follow these steps:

Step 1: Express $f(x + h)$

$f(x + h) = \sqrt{17(x + h)} = \sqrt{17x + 17h}$

Step 2: Write the difference quotient formula

The difference quotient is given by: $\frac{f(x + h) - f(x)}{h} = \frac{\sqrt{17x + 17h} - \sqrt{17x}}{h}$

Step 3: Simplify the expression

To simplify the expression, we will rationalize the numerator by multiplying both the numerator and denominator by the conjugate of the numerator, which is $\sqrt{17x + 17h} + \sqrt{17x}$.

$\frac{\sqrt{17x + 17h} - \sqrt{17x}}{h} \times \frac{\sqrt{17x + 17h} + \sqrt{17x}}{\sqrt{17x + 17h} + \sqrt{17x}} = \frac{\left( \sqrt{17x + 17h} \right)^2 - \left( \sqrt{17x} \right)^2}{h \left( \sqrt{17x + 17h} + \sqrt{17x} \right)}$

Step 4: Simplify the numerator

Using the difference of squares formula: $\left( \sqrt{17x + 17h} \right)^2 - \left( \sqrt{17x} \right)^2 = (17x + 17h) - 17x = 17h$

Now, the expression becomes: $\frac{17h}{h \left( \sqrt{17x + 17h} + \sqrt{17x} \right)}$

Step 5: Cancel out $h$

Since $h \neq 0$, we can cancel $h$ from the numerator and denominator: $\frac{17}{\sqrt{17x + 17h} + \sqrt{17x}}$

This is the simplified form of the difference quotient.

Final Answer:

$\frac{17}{\sqrt{17x + 17h} + \sqrt{17x}}$

Would you like more details or have any questions?

Here are 5 questions to explore related concepts:

1. How does the difference quotient relate to the derivative of a function?
2. What is the difference quotient of $f(x) = x^2$?
3. How does the conjugate help in simplifying the expression?
4. Can you find the derivative of $f(x) = \sqrt{17x}$ using the difference quotient definition?
5. What are the limitations of using the difference quotient for non-differentiable functions?

Tip: Rationalizing the numerator is a common technique for simplifying expressions involving square roots!