zero de funció

Si f(x) = (x+(-a))+(x+(-b)) ==> [∃c][ c€[a+s,b+(-s)]_{K} & f(c) = 0 ]
f(a+s) = (-1)·f(b+(-s))
f(b+(-s)) = (-1)·f(a+s)


Si s = ( (b+(-a))/2 ) ==> f( (a+b)/2 ) = 0


Si f(x) = ( s/(x+(-a)) )+( s/(x+(-b)) ) ==> [∃c][ c€[a+s,b+(-s)]_{K} & f(c) = 0 ]
f(a+s) = (-1)·f(b+(-s))
f(b+(-s)) = (-1)·f(a+s)


Si s = ( (b+(-a))/2 ) ==> f( (a+b)/2 ) = 0


Si f(x) = (x+(-1)·a_{1})+...(n)...+(x+(-1)·a_{n}) ==> f( (a_{1}+...(n)...+a_{n})/n ) = 0


0 = nx+(-1)( a_{1}+...(n)...+a_{n} )
a_{1}+...(n)...+a_{n} = nx