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]?

Question

Scarlett Allison · Accepted Answer

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]#.

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).

Isaiah Allen · Answer

undefined To use the Intermediate Value Theorem to show that the polynomial function $x^3 + 2x^2 - 42$ has a root in the interval $[0, 3]$, you need to demonstrate that the function changes sign over that interval. First, evaluate the function at the endpoints of the interval: $f(0)$ and $f(3)$. If the values have different signs, then there exists at least one root of the function in the interval $[0, 3]$.

$f(0) = 0^3 + 2(0)^2 - 42 = -42$

$f(3) = 3^3 + 2(3)^2 - 42 = 27 + 18 - 42 = 3$

Since $f(0) = -42$ and $f(3) = 3$, and $f(0)$ is negative while $f(3)$ is positive, the function changes sign over the interval $[0, 3]$. Thus, by the Intermediate Value Theorem, there exists at least one root of the function $x^3 + 2x^2 - 42$ in the interval $[0, 3]$.