8A F11 Q12

Question: Find and simplify the difference quotient ${\frac {f(x+h)-f(x)}{h}}$ for f(x) = ${\frac {2}{3x+1}}$

Foundations
1) f(x + h) = ?
2) How do you eliminate the 'h' in the denominator?
Answer:
1) Since $f(x+h)={\frac {2}{3(x+h)+1}}$ the difference quotient is a difference of fractions divided by h.
2) The numerator is ${\frac {2}{3(x+h)+1}}-{\frac {2}{3x+1}}$ so the first step is to simplify this expression. This then allows us to eliminate the 'h' in the denominator.

Solution:

Step 1:
The difference quotient that we want to simplify is ${\frac {f(x+h)-f(x)}{h}}=\left({\frac {2}{3(x+h)+1}}-{\frac {2}{3x+1}}\right)\div h$

Step 2:
Now we simplify the numerator:
${\begin{array}{rcl}{\frac {f(x+h)-f(x)}{h}}&=&\left({\frac {2}{3(x+h)+1}}-{\frac {2}{3x+1}}\right)\div h\\&=&{\frac {2(3x+1)-2(3(x+h)+1)}{h(3(x+h)+1)(3x+1))}}\end{array}}$

Arithmetic:
Now we simplify the numerator:
${\begin{array}{rcl}{\frac {2(3x+1)-2(3(x+h)+1)}{h(3(x+h)+1)(3x+1))}}&=&{\frac {6x+2-6x-6h-2}{h(3(x+h)+1)(3x+1))}}\\&=&{\frac {-6}{(3(x+h)+1)(3x+1))}}\end{array}}$

Final Answer:
${\frac {-6}{(3(x+h)+1)(3x+1))}}$

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