Questions: Let the function f be defined as follows. f(x) = 6 sqrt(x-3) Answer the following. (a) For h != 0, find the difference quotient (f(x+h)-f(x)) / h. Simplify your answer so that the resulting expression is defined at h=0. (f(x+h)-f(x)) / h =

Solution Steps

To find the difference quotient \(\frac{f(x+h)-f(x)}{h}\) for the function \(f(x) = 6 \sqrt{x-3}\), we need to:

1. Substitute \(f(x)\) and \(f(x+h)\) into the difference quotient formula.
2. Simplify the expression to make it defined at \(h=0\).

Given the function \( f(x) = 6 \sqrt{x-3} \), we need to find the difference quotient: \[ \frac{f(x+h) - f(x)}{h} \]

First, we substitute \( f(x) \) and \( f(x+h) \) into the difference quotient: \[ f(x) = 6 \sqrt{x-3} \] \[ f(x+h) = 6 \sqrt{x+h-3} \] \[ \frac{f(x+h) - f(x)}{h} = \frac{6 \sqrt{x+h-3} - 6 \sqrt{x-3}}{h} \]

We can factor out the common factor of 6 in the numerator: \[ \frac{6 \sqrt{x+h-3} - 6 \sqrt{x-3}}{h} = 6 \cdot \frac{\sqrt{x+h-3} - \sqrt{x-3}}{h} \]

To simplify further, we rationalize the numerator by multiplying and dividing by the conjugate: \[ 6 \cdot \frac{\sqrt{x+h-3} - \sqrt{x-3}}{h} \cdot \frac{\sqrt{x+h-3} + \sqrt{x-3}}{\sqrt{x+h-3} + \sqrt{x-3}} \] \[ = 6 \cdot \frac{(\sqrt{x+h-3})^2 - (\sqrt{x-3})^2}{h (\sqrt{x+h-3} + \sqrt{x-3})} \] \[ = 6 \cdot \frac{(x+h-3) - (x-3)}{h (\sqrt{x+h-3} + \sqrt{x-3})} \] \[ = 6 \cdot \frac{h}{h (\sqrt{x+h-3} + \sqrt{x-3})} \] \[ = 6 \cdot \frac{1}{\sqrt{x+h-3} + \sqrt{x-3}} \]

Final Answer