8A F11 Q12

Question: Find and simplify the difference quotient Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{f(x+h)-f(x)}{h}} for f(x) = Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{2}{3x+1}}

Foundations
1) f(x + h) = ?
2) How do you eliminate the 'h' in the denominator?
Answer:
1) Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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:
${\begin{array}{rcl}{\frac {f(x+h)-f(x)}{h}}&=&\left({\frac {2}{3(x+h)+1}}-{\frac {2}{3x+1}}\right)\div h\\&=&{\frac {2(3x+1)-2(3(x+h)+1)}{h(3(x+h)+1)(3x+1))}}\end{array}}$