How do I use the intermediate value theorem to prove every polynomial of odd degree has at least one real root?

1 Answer
Oct 27, 2015

Answer:

Given any polynomial #f(x)# of odd degree and positive leading coefficient find #x_1# such that #f(-x_1) < 0# and #f(x_1) > 0#, so #EE x in (-x_1, x_1)# with #f(x) = 0#.

Explanation:

Let #f(x) = a_0x^n+a_1x^(n-1)+...+a_n# with #a_0 != 0#

Note that #f(x)# is a continuous function.

If #x# is sufficiently large and positive, #f(x) > 0#.

To prove that:

Let #x_1 = (1+abs(a_0)+abs(a_1)+abs(a_2)+...+abs(a_n))/abs(a_0)#

Note that #x_1 > 1#, so if #m <= n-1# then #x^m <= x^(n-1)#

So we find:

#f(x_1) >= a_0x_1^n-(abs(a_1)x^(n-1)+abs(a_2)x^(n-2)+...+abs(a_n))#

#>= a_0x_1^n-(abs(a_1)x^(n-1)+abs(a_2)x^(n-1)+...+abs(a_n)x^(x-1))#

#= x^(n-1)(a_0x-(abs(a_1)+abs(a_2)+...+abs(a_n)))#

#= x^(n-1)((1+abs(a_0)+abs(a_1)+abs(a_2)+...+abs(a_n))-(abs(a_1)+abs(a_2)+...+abs(a_n)))#

#= x^(n-1)(1+abs(a_0)) > 0#

If #n# is odd and #x# sufficiently large and negative then #f(x) < 0#.

To prove that, note that if #n# is odd then #a_0(-x)^n = -a_0x^n#, so #-f(-x) = a_0x^n-a_1x^(n-1)+...-a_n#, hence #f(-x_1) < 0#

We have:

  • #f(x)# continuous over #[-x_1, x_1]#
  • #f(-x_1) < 0 < f(x_1)#

So by the intermediate value theorem:

#EE x in (-x_1, x_1) : f(x) = 0#

If the leading coefficient (#a_0#) of #f(x)# is negative then the same formula for #x_1# yields #f(x_1) < 0 < f(-x_1)#, hence there is a root in #(-x_1, x_1)# again.