Question 1121642

even tests are{{{ f(-x) = f(x)}}}, 

given: {{{f(x)=(x^2+1)/x }}}

{{{f(-x)=((-x)^2+1)/(-x)}}}

{{{f(-x) = -(x^2+1)/x }}}=> {{{ f(-x) = -f(x)}}}

so,  {{{f(-x) <> f(x)  }}}=> {{{f(x)=(x^2+1)/x}}}  is  {{{not}}} an even function


odd tests are {{{f(-x) = -f(x)}}}

since above is proven that f(-x) = -f(x), means f(x) = (x^2 + 1)/x is an {{{odd}}} function


Since the function is {{{ odd}}}, it is  symmetric about the {{{origin}}}.


{{{ graph( 600, 600, -10, 10, -10, 10, (x^2+1)/x) }}}