Question

standard 1 homework
score: 25.5/28 answered: 19/20
question 20
determine and fully simplify the difference quotient, (\frac{f(a + h) - f(a)}{h}), for the function (f(x) = \frac{1}{9x}).
(f(a + h) = square)
(f(a) = square)
(\frac{f(a + h) - f(a)}{h} = square)
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Explanation:

Step1: Find \( f(a + h) \)

Substitute \( x = a + h \) into \( f(x)=\frac{1}{9x} \).
\( f(a + h)=\frac{1}{9(a + h)} \)

Step2: Find \( f(a) \)

Substitute \( x = a \) into \( f(x)=\frac{1}{9x} \).
\( f(a)=\frac{1}{9a} \)

Step3: Compute \( \frac{f(a + h)-f(a)}{h} \)

First, find \( f(a + h)-f(a) \):
\( f(a + h)-f(a)=\frac{1}{9(a + h)}-\frac{1}{9a} \)
Find a common denominator, which is \( 9a(a + h) \):
\(=\frac{a}{9a(a + h)}-\frac{a + h}{9a(a + h)}=\frac{a-(a + h)}{9a(a + h)}=\frac{a - a - h}{9a(a + h)}=\frac{-h}{9a(a + h)} \)
Then divide by \( h \):
\( \frac{f(a + h)-f(a)}{h}=\frac{\frac{-h}{9a(a + h)}}{h} \)
Divide the fractions: \( \frac{-h}{9a(a + h)}\times\frac{1}{h}=-\frac{1}{9a(a + h)} \) (assuming \( h
eq0 \))

Answer:

\( f(a + h)=\frac{1}{9(a + h)} \)
\( f(a)=\frac{1}{9a} \)
\( \frac{f(a + h)-f(a)}{h}=-\frac{1}{9a(a + h)} \)