Question 1196650
{{{f(x) =(-1)}}}^[x] is even, odd or neither 

Let {{{f}}}:{{{R}}}→{{{R}}} be a function defined as {{{f(x)=(-1)}}}^[x] 
 .
When {{{x }}}is an integer [x]={{{x}}}, hence  {{{f(x)=(-1)}}}^[x] ={{{(-1)^x}}}, which is an even function.


If {{{x}}} is a positive real number which is not an integer, then [x] is the integer part of the given number, say {{{a}}}.

Therefore  {{{f(x)}}}=±{{{1}}} based on {{{a}}} is even or odd.

If {{{x }}}is a negative real number which is not an integer, then [x]={{{a^-1}}}, where {{{a}}} is the integer part of the given number.

Therefore  {{{f(x)}}}=±{{{1}}} based on {{{a}}} is {{{odd}}} or {{{even}}}.

Hence {{{f(x)=-f(-x)}}} when {{{x}}} not  ∈{{{Z }}}
 

so, correct answers are:
{{{f(x)}}}  is an {{{even}}} function when{{{ x}}} ∈{{{ Z}}}
{{{f(x)}}}  is an {{{odd }}}function when {{{x}}} is {{{not}}} ∈{{{Z}}}
{{{f(x)}}}  is {{{neither }}}even nor odd