#### Answer

a)$\frac{-2h}{h+1}$
b) $\frac{-2h^2}{h+1}$

#### Work Step by Step

a) $g(0)=\frac{2}{0+1}=2$
$g(h)=\frac{2}{h+1}$
$\frac{2}{h+1}-2=\frac{-2h}{h+1}$
b) $\frac{\frac{-2h}{h+1}}{h-0}$
= ${\frac{-2h}{h+1}}\times\frac{1}{h}$
=$\frac{-2}{h+1}$

Published by
Brooks Cole

ISBN 10:
1305071751

ISBN 13:
978-1-30507-175-9

a)$\frac{-2h}{h+1}$
b) $\frac{-2h^2}{h+1}$

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