## Algebra 2 (1st Edition)

Published by McDougal Littell

# Chapter 7 Exponential and Logarithmic Functions - Prerequisite Skills - Page 476: 9

#### Answer

$$y=\sqrt{2}\sqrt{x},\:y=-\sqrt{2}\sqrt{x}$$

#### Work Step by Step

Switching the elements of the domain and the range (in other words, the dependent and independent variables), we find that the inverse is: $$x=\frac{1}{2}y^2 \\ y=\sqrt{2}\sqrt{x},\:y=-\sqrt{2}\sqrt{x}$$

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