Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter P - Section P.6 - Rational Expressions - Exercise Set - Page 85: 71


The required solution $-\frac{2x+h}{{{x}^{2}}{{\left( x+h \right)}^{2}}}$.

Work Step by Step

We have the given complex rational expression $\frac{\frac{1}{{{\left( x+h \right)}^{2}}}-\frac{1}{{{x}^{2}}}}{h}$. We know that a rational expression is the one that can be expressed in the form $\frac{p}{q}$, where, both $p\ \text{and }q$ are polynomials and the denominator, $q\ne 0$. Furthermore, a complex rational expression or a complex fraction is an algebraic rational expression in which either the numerator or denominator or both the numerator and denominator contain rational expressions. And the given complex rational expression is $\frac{\frac{1}{{{\left( x+h \right)}^{2}}}-\frac{1}{{{x}^{2}}}}{h}$. The numerator is $\frac{1}{{{\left( x+h \right)}^{2}}}-\frac{1}{{{x}^{2}}}$. Simplify the numerator: $\begin{align} & \frac{1}{{{\left( x+h \right)}^{2}}}-\frac{1}{{{x}^{2}}}=\frac{1}{{{\left( x+h \right)}^{2}}}\times \frac{{{x}^{2}}}{{{x}^{2}}}-\frac{1}{{{x}^{2}}}\times \frac{{{\left( x+h \right)}^{2}}}{{{\left( x+h \right)}^{2}}} \\ & =\frac{{{x}^{2}}}{{{x}^{2}}{{\left( x+h \right)}^{2}}}-\frac{{{\left( x+h \right)}^{2}}}{{{x}^{2}}{{\left( x+h \right)}^{2}}} \\ & =\frac{{{x}^{2}}-{{\left( x+h \right)}^{2}}}{{{x}^{2}}{{\left( x+h \right)}^{2}}} \\ & =\frac{{{x}^{2}}-\left( {{x}^{2}}+2xh+{{h}^{2}} \right)}{{{x}^{2}}{{\left( x+h \right)}^{2}}} \end{align}$. Simplifying further, $\begin{align} & \frac{1}{{{\left( x+h \right)}^{2}}}-\frac{1}{{{x}^{2}}}=\frac{{{x}^{2}}-{{x}^{2}}+2xh-{{h}^{2}}}{{{x}^{2}}{{\left( x+h \right)}^{2}}} \\ & =\frac{2xh-{{h}^{2}}}{{{x}^{2}}{{\left( x+h \right)}^{2}}}. \end{align}$ And simplify the given complex rational expression: $\begin{align} & \frac{\frac{1}{{{\left( x+h \right)}^{2}}}-\frac{1}{{{x}^{2}}}}{h}=\frac{\frac{-2xh-{{h}^{2}}}{{{x}^{2}}{{\left( x+h \right)}^{2}}}}{h} \\ & =\frac{-h\left( 2x+h \right)}{{{x}^{2}}{{\left( x+h \right)}^{2}}}\cdot \frac{1}{h} \\ & =-\frac{2x+h}{{{x}^{2}}{{\left( x+h \right)}^{2}}} \end{align}$. Hence, $\frac{\frac{1}{{{\left( x+h \right)}^{2}}}-\frac{1}{{{x}^{2}}}}{h}=$ $-\frac{2x+h}{{{x}^{2}}{{\left( x+h \right)}^{2}}}$ ; $x\ne 0\ ;\ h\ne 0\ ;\ x\ne h$.
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