Question

find $f(a), f(a + h)$, and the difference quotient $\frac{f(a + h)-f(a)}{h}$, where $h
eq0$. $f(x)=\frac{3x}{x - 8}$. $f(a)=\frac{3a}{a - 8}$, $f(a + h)=\frac{3a + 3h}{a + h - 8}$, $\frac{f(a + h)-f(a)}{h}=$

Explanation:

Step1: Substitute \(f(a + h)\) and \(f(a)\) into the difference - quotient formula

We know \(f(a + h)=\frac{3a + 3h}{a + h-8}\) and \(f(a)=\frac{3a}{a - 8}\). The difference quotient \(\frac{f(a + h)-f(a)}{h}=\frac{\frac{3a + 3h}{a + h-8}-\frac{3a}{a - 8}}{h}\).

Step2: Find a common denominator for the numerator

The common denominator of \(a + h-8\) and \(a - 8\) is \((a + h-8)(a - 8)\). So \(\frac{3a + 3h}{a + h-8}-\frac{3a}{a - 8}=\frac{(3a + 3h)(a - 8)-3a(a + h-8)}{(a + h-8)(a - 8)}\).
Expand the numerator: \((3a + 3h)(a - 8)=3a^{2}-24a+3ah - 24h\) and \(3a(a + h-8)=3a^{2}+3ah-24a\). Then \((3a + 3h)(a - 8)-3a(a + h-8)=3a^{2}-24a+3ah - 24h-(3a^{2}+3ah-24a)=-24h\).
So \(\frac{(3a + 3h)(a - 8)-3a(a + h-8)}{(a + h-8)(a - 8)}=\frac{-24h}{(a + h-8)(a - 8)}\).

Step3: Divide by \(h\)

\(\frac{\frac{-24h}{(a + h-8)(a - 8)}}{h}=\frac{-24}{(a + h-8)(a - 8)}\).

Answer:

\(\frac{-24}{(a + h-8)(a - 8)}\)