Answer
a) $\dfrac{-2h}{a(a+h)}$
b) $\dfrac{-2}{a(a+h)}$
Work Step by Step
a) Apply the rule such as: $f(b)-f(a)$ where $a \leq by$
Thus, the net change is given as:
$f(b)-f(a)=f(a+h)-f(h)$
This gives:
$\dfrac{2}{a+h}-\dfrac{2}{a}=\dfrac{2a-2a-2h}{a(a+h)}=\dfrac{-2h}{a(a+h)}$
b) Apply the rule such as: $f(b)-f(a)$ where $a \leq by$
Thus, the net change is given as:
$\dfrac{f(b)-f(a)}{b-a}=\dfrac{f(a+h)-f(a)}{(a+h)-a}$
This gives:
$\dfrac{2/a+h-2/a}{a+h-a}=\dfrac{\dfrac{-2h}{a(a+h)}}{h}=\dfrac{-2}{a(a+h)}$
