# 008A Sample Final A, Question 12

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

Foundations
1) f(x + h) = ?
2) How do you eliminate the 'h' in the denominator?
1) Since ${\displaystyle f(x+h)={\frac {2}{3(x+h)+1}}}$ the difference quotient is a difference of fractions divided by h.
2) The numerator is ${\displaystyle {\frac {2}{3(x+h)+1}}-{\frac {2}{3x+1}}}$ 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 ${\displaystyle {\frac {f(x+h)-f(x)}{h}}=\left({\frac {2}{3(x+h)+1}}-{\frac {2}{3x+1}}\right)\div h}$
Step 2:
Now we simplify the numerator:

${\displaystyle {\begin{array}{rcl}{\frac {f(x+h)-f(x)}{h}}&=&\left({\frac {2}{3(x+h)+1}}-{\frac {2}{3x+1}}\right)\div h\\&=&{\frac {2(3x+1)-2(3(x+h)+1)}{h(3(x+h)+1)(3x+1))}}\end{array}}}$

Arithmetic:
Now we simplify the numerator:

${\displaystyle {\begin{array}{rcl}{\frac {2(3x+1)-2(3(x+h)+1)}{h(3(x+h)+1)(3x+1))}}&=&{\frac {6x+2-6x-6h-2}{h(3(x+h)+1)(3x+1))}}\\&=&{\frac {-6}{(3(x+h)+1)(3x+1))}}\end{array}}}$

${\displaystyle {\frac {-6}{(3(x+h)+1)(3x+1))}}}$