$$\text { Find and simplify the difference quotient } \frac { f ( a + h ) - f ( a ) } { h } , h \neq 0 \text { for the given function. }$$
-$f ( x ) = \frac { 2 } { x + 1 }$
A) $\frac { - 2 } { a ( a + h + 1 ) }$
B) $\frac { - 2 } { ( a + 1 ) ( a + h + 1 ) }$
C) $\frac { - 2 } { ( a + 1 ) ( a + h ) }$
D) $\frac { - 2 } { a ( a + h ) }$

Question

Quizplus · Accepted Answer

To find the difference quotient for $f(x) = \frac{2}{x+1}$, we substitute $x$ with $a$ and $a+h$ into the function, then simplify the expression $\frac{f(a+h)-f(a)}{h}$. First, find $f(a+h)$ and $f(a)$:- $f(a+h) = \frac{2}{(a+h)+1} = \frac{2}{a+h+1}$- $f(a) = \frac{2}{a+1}$Now, compute the difference quotient:$$\frac{f(a+h)-f(a)}{h} = \frac{\frac{2}{a+h+1} - \frac{2}{a+1}}{h}$$To simplify, find a common denominator for the fractions in the numerator:$$= \frac{\frac{2(a+1) - 2(a+h+1)}{(a+h+1)(a+1)}}{h}$$$$= \frac{2(a+1) - 2(a+h+1)}{h(a+h+1)(a+1)}$$$$= \frac{2a+2 - 2a - 2h - 2}{h(a+h+1)(a+1)}$$$$= \frac{-2h}{h(a+h+1)(a+1)}$$$$= \frac{-2}{(a+h+1)(a+1)}$$Therefore, the simplified difference quotient is $\frac{-2}{(a+h+1)(a+1)}$, which matches choice B.