Question

Is f(x)=x2+sinx an even or odd function?

Answer 1

$if \\ \\ f(x) = f( - x) \\ \\ f(x) \: is \: an \: even \: function \\ \\ if \\ \\ f(x) = - f( - x) \: \\ then \: it \: is \: an \: odd \: function \\ \\ if \\ \\ f(x) \: not \: equals \: to \: f( - x) \: or \: - f( - x) \: \\ then \: f(x) \: is \: niether \: odd \: nor \: even \: function$

Now

$f(x) = {x}^{2} + \sin \: x \\ \\ f( - x) = {( - x)}^{2} + \sin( - x ) \\ \\ f( - x) = {x}^{2} - \sin \: x$

Therefore f(x) is neither odd nor even function.

hope it helps

Answer 2

(x)=f(−x)f(x)isanevenfunctioniff(x)=−f(−x)thenitisanoddfunctioniff(x)notequalstof(−x)or−f(−x)thenf(x)isnietheroddnorevenfunction​

Now

$\begin{lgathered}f(x) = {x}^{2} + \sin \: x$\\ \\ f( - x) = {( - x)}^{2} + \sin( - x ) \\ \\ f( - x) = {x}^{2} - \sin \: x\end{lgathered}f(x)=x2+sinxf(−x)=(−x)2+sin(−x)f(−x)=x2−sinx​

Therefore f(x) is neither odd nor even function.

hope it helps