Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Review - Exercises: 50

Answer

$\dfrac{\sqrt{x+h}-\sqrt{x}}{h}=\dfrac{1}{\sqrt{x+h}+\sqrt{x}}$

Work Step by Step

$\dfrac{\sqrt{x+h}-\sqrt{x}}{h}$ (rationalize the numerator) Multiply the numerator and the denominator of the rational expression by the conjugate of the numerator and simplify: $\dfrac{\sqrt{x+h}-\sqrt{x}}{h}=\dfrac{\sqrt{x+h}-\sqrt{x}}{h}\cdot\dfrac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}=...$ $...=\dfrac{(\sqrt{x+h})^{2}-(\sqrt{x})^{2}}{h(\sqrt{x+h}+\sqrt{x})}=\dfrac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}=...$ $...=\dfrac{h}{h(\sqrt{x+h}+\sqrt{x})}=\dfrac{1}{\sqrt{x+h}+\sqrt{x}}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.