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

$\mathrm{Even}.$
$\mathrm{Function\:Parity\:Definition:}$ $\mathrm{Even\:Function:}\:\:$ A function is even if $\ g(-x)=g(x)\$ for all $\ x\in \mathbb{R}.$ $\mathrm{Odd\:Function:}\:\:$ A function is odd if $\ g(-x)=-g(x)\$ for all $\ x\in \mathbb{R}.$ $g(x)=\frac{1}{x^2-1}$ $g(-x)=\frac{1}{(-x)^2-1}$ $g(-x)=\frac{1}{x^2-1}$ Now, $-g(x)=-\frac{1}{x^2-1}$ Since, $g(-x)=g(x)\mathrm{,\:therefore\:}\frac{1}{x^2-1}\mathrm{\:is\:an\:even\:function}$ $g(-x)\ne-g(x)\mathrm{,\:therefore\:}\frac{1}{x^2-1}\mathrm{\:is\:not\:an\:odd\:function}$