Big Ideas Math - Algebra 1, A Common Core Curriculum

Published by Big Ideas Learning LLC
ISBN 10: 978-1-60840-838-2
ISBN 13: 978-1-60840-838-2

Chapter 10 - Radical Functions and Equations - 10.4 - Inverse of a Function - Exercises - Page 572: 13

Answer

$x=\pm\frac{\sqrt y}{3}$ When output $y=2$, input $x=\pm\frac{\sqrt 2}{3}$

Work Step by Step

$y=9x^{2}$ Divide both sides of the equation by $9$ to get $\frac{y}{9}=x^{2}$ Taking square root on both sides, we have $\pm\frac{\sqrt y}{3}=x$ That is, $x=\pm\frac{\sqrt y}{3}$ When output $y=2$, Input $x=\pm\frac{\sqrt 2}{3}$
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