Precalculus: Mathematics for Calculus, 7th Edition

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

Chapter 1 - Section 1.11 - Solving Equations and Inequalities Graphically - 1.11 Exercises - Page 121: 10

Answer

$x$ $\approx$ $-2.52$

Work Step by Step

Given equation is- $x^{3}+16$ = $0$ Algebraic Solution: $x^{3}$ = $-16$ i.e. $x^{3}$ = $(-4\sqrt 2)^{2}$ i.e. $x$ = $\sqrt[3] {-16}$ i.e. $x$ = $-2.5198420998$ i.e. $x$ $\approx$ $-2.52$ Graphical Solution: $x^{3}+16$ = $0$ Now graphing $y$ = $x^{3}+16$ $x-intercepts$ is $-2.5198$ i.e. $x$ = $-2.5198$ i.e. $x$ $\approx$ $-2.52$
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