## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 1 - Section 1.1 - Functions and Their Graphs - Exercises - Page 12: 38

#### Answer

Graph is symmetric with respect to $\ \mathrm{y-axis}.$ The graph of the given function $\ y=-\frac{1}{x^2}\$ is decreasing on $\ (-\infty,0)\$ and increasing on $\ (0,\infty).$ #### Work Step by Step

$\mathrm{First\:Part:}\:\:$ According to the definitions: A function $\ f\$ defined on an interval is increasing on $\ (a, b)\$ if for every $\ x_1, x_2\$ $\in$ $(a, b)$ $\ x_1\le x_2\$ implies that $\ f(x_1)\le f(x_2).\$ A function $\ f\$ defined on an interval is decreasing on $\ (a, b)\$ if for every $\ x_1, x_2\$ $\in$ $(a, b)$ $\ x_1\le x_2\$ implies that $\ f(x_1)\ge f(x_2).\$ First of all create a table with a few points to sketch the graph. $\quad \mathrm{See\:the\:table\:and\:graph\:above.}$ $\mathrm{Second\:Part:}\:\:$ Graph is symmetric with respect to $\ \mathrm{y-axis}.$ $\mathrm{Third\:Part:}\:\:$ The graph of the given function $\ y=-\frac{1}{x^2}\$ is decreasing on $\ (-\infty,0)\$ and increasing on $\ (0,\infty).$

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