Answer
increasing on $(0,\infty)$, decreasing on $(-\infty,0)$.
Work Step by Step
Function $f(x)=x^2$ opens up with a vertex at $(0,0)$, thus it is increasing on $(0,\infty)$, decreasing on $(-\infty,0)$.
Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)
by Sullivan III, Michael
Published by
Pearson
ISBN 10:
0-32193-104-1
ISBN 13:
978-0-32193-104-7
Chapter 4 - Exponential and Logarithmic Functions - Section 4.2 One-to-One Functions; Inverse Functions - 4.2 Assess Your Understanding - Page 290: 2
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