#### Answer

$h'(x)=1$

#### Work Step by Step

Simplify the expression:
$h(x)=\dfrac{x^3-6x^2+8x}{x^2-2x}=\dfrac{x(x^2-6x+8)}{x(x-2)}$
$h(x)=\dfrac{x^2-6x+8}{x-2}=\dfrac{(x-4)(x-2)}{x-2}$
$h(x)=x+4$
Apply power rule
$h'(x)=1$

Published by
Pearson

ISBN 10:
0321947347

ISBN 13:
978-0-32194-734-5

$h'(x)=1$

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