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: 77


The simplified form of the expression $\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}}{h}$ is $\frac{\sqrt{x}-\sqrt{x+h}}{h\sqrt{x\left( x+h \right)}}$.

Work Step by Step

Consider the provided expression, $\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}}{h}$ Multiply the numerator and denominator by $\sqrt{x}\sqrt{x+h}$. Therefore, $\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}}{h}=\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}}{h}\cdot \frac{\sqrt{x}\sqrt{x+h}}{\sqrt{x}\sqrt{x+h}}$ Apply the distributive property in the numerator: $a\left( b+c \right)=ab+ac$ Therefore, $$ $\begin{align} & \frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}}{h}=\frac{\frac{1}{\sqrt{x+h}}\sqrt{x}\sqrt{x+h}-\frac{1}{\sqrt{x}}\sqrt{x}\sqrt{x+h}}{h\sqrt{x}\sqrt{x+h}} \\ & =\frac{\sqrt{x}-\sqrt{x+h}}{h\sqrt{x}\sqrt{x+h}} \\ & =\frac{\sqrt{x}+\sqrt{x+h}}{h\sqrt{x\left( x+h \right)}} \end{align}$ $$ Here, $h\ne 0$ Therefore, the simplified form of the expression $\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}}{h}$ is $\frac{\sqrt{x}-\sqrt{x+h}}{h\sqrt{x\left( x+h \right)}}$.
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