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I know IVT can be used to prove a function has at least 1 root but can one use squeeze thm? I had a student try to prove:

$$x^3-cosx$$

has at least one root using squeeze thm but this is definitely incorrect:

$$x^3-1<=x^3-cosx<=x^3+1$$

For this (or in general) is it possible to prove a root exists using squeeze thm?

1 Answers 1

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This approach actually technically works, but it requires more details - and it isn't actually using the Squeeze Theorem. The Theorem states that if $g(x) < f(x) < h(x)$ and $\lim_{x\to c}g(x) = \lim_{x\to c}h(x)$, then $\lim_{x\to c}f(x) = \lim_{x\to c}g(x)$. Since no limits are involved in what you outlined, the Squeeze Theorem isn't relevant.

However, it is the case that if $g(x) \leq f(x) \leq h(x)$ and both $g$ and $h$ have at least one root, then $f$ has at least one root. This is because at the root $c$ of $h$, $h(c) = 0 \geq f(c) \geq g(c)$, while at the root $b$ of $g$ we have $g(b) = 0 \leq f(b) \leq h(b)$. So at $c$, $f$ is at most zero; at $b$, $f$ is at least zero. By the IVT, $f$ has a root somewhere between $c$ and $b$.

I'm not sure I'd give a student credit, though, unless they actually explained this whole thing - to my knowledge, this is not a standard theorem so much as a disguised use of the IVT.

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    You're right since they didn't use limits the squeeze theorem wouldn't apply, plus I don't think there is a c that satisfies the limit condition. Technically they used your second example without full explanation, I guess I shouldn't give them the marks (3 of my top students used this method). – 2017-02-17