Question

find the difference quotient of f; that is find \\(\frac{f(x + h) - f(x)}{h}\\), \\(h \
eq 0\\), for the function \\(f(x) = \sqrt{x - 6}\\). hint: rationalize the numerator. \\(\dots\\) the difference quotient of f; \\(f(x) = \sqrt{x - 6}\\) is \\(\square\\). (simplify your answer.)

Explanation:

Step1: Substitute \( f(x+h) \) and \( f(x) \)

First, find \( f(x + h) \) for \( f(x)=\sqrt{x - 6} \). So \( f(x + h)=\sqrt{(x + h)-6}=\sqrt{x + h - 6} \). Then substitute into the difference quotient formula:
\[
\frac{f(x + h)-f(x)}{h}=\frac{\sqrt{x + h - 6}-\sqrt{x - 6}}{h}
\]

Step2: Rationalize the numerator

Multiply the numerator and denominator by the conjugate of the numerator, which is \( \sqrt{x + h - 6}+\sqrt{x - 6} \):
\[
\frac{(\sqrt{x + h - 6}-\sqrt{x - 6})(\sqrt{x + h - 6}+\sqrt{x - 6})}{h(\sqrt{x + h - 6}+\sqrt{x - 6})}
\]
Using the difference of squares formula \( (a - b)(a + b)=a^2 - b^2 \) in the numerator:
\[
\frac{(\sqrt{x + h - 6})^2-(\sqrt{x - 6})^2}{h(\sqrt{x + h - 6}+\sqrt{x - 6})}=\frac{(x + h - 6)-(x - 6)}{h(\sqrt{x + h - 6}+\sqrt{x - 6})}
\]

Step3: Simplify the numerator

Simplify the numerator: \( (x + h - 6)-(x - 6)=x + h - 6 - x + 6 = h \)
So now the expression becomes:
\[
\frac{h}{h(\sqrt{x + h - 6}+\sqrt{x - 6})}
\]

Step4: Cancel out \( h \)

Since \( h
eq0 \), we can cancel the \( h \) in the numerator and denominator:
\[
\frac{1}{\sqrt{x + h - 6}+\sqrt{x - 6}}
\]

Answer:

\(\frac{1}{\sqrt{x + h - 6}+\sqrt{x - 6}}\)