Question

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]?

Answer 1

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).