Question

Question #c3696

Answer 1

Intermediate Value Theorem

If a function `f` is continuous on `[a,b]`, and `N` is any number between `f(a)` and `f(b)`, then there exists some number `c` in `(a,b)` such that `f(c)=N`.

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Let us prove that `x^3-x-1=0` has at least one root.

Proof

Let `f(x)=x^3-x-1`, and let `N=0`.

Since `f(0)=-1` and `f(2)=5`, `f(0) < 0 < f(2)`.

`f` is continuous on `[0,2]` since it is a polynomial function.

By Intermediate Value Theorem, there exists some number `c` in `(0,2)` such that `f(c)=c^3-c-1=0`.

Hence, the equation has at least one root `c`.

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I hope that this was helpful.