How do you use the Intermediate Value Theorem to show that the polynomial function # x^3+2x^2-42# has a root in the interval [0, 3]?

1 Answer
George C.
Oct 9, 2015

This function is negative at #0#, positive at #3# and continuous on the whole interval #[0, 3]#. So it must take the value #0# for some #x in [0, 3]#.

Explanation:

Let #f(x) = x^3+2x^2-42#.

Then #f(x)# is continuous at every #x in RR#.

#f(0) = 0 + 0 - 42 = -42 < 0#

#f(3) = 27 + 18 - 42 = 3 > 0#

So by the Intermediate Value Theorem, #EE x in [0, 3] : f(x) = 0#

The Intermediate Value Theorem is one way of formulating the completeness of #RR# as an ordered field (a set of numbers closed under addition, multiplication and their inverses, etc).

Notice that the rational numbers #QQ# do not satisfy the Intermediate Value Theorem. For example, the function #g(x) = x^3 - 2# satisfies #g(0) = -2 < 0# and #g(2) = 6 > 0#, but there is no #x in QQ nn [0, 2]# that satisfies #g(x) = 0# (since #root(3)(2)# is irrational).

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