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# Existence of roots

Let be a continuous function such that

(1)

Prove that there exist such that and .

Solution

The existence of follows from Mean Value Theorem of Integrals. To prove that we proceed as follows:

We are given that  . Suppose f is not identically zero for otherwise the result is trivial. The condition implies there is at least one sign-changing root, say at (that is, the function has different signs after passing through , or more formally, for all sufficiently small ) Suppose this is the only sign-changing root. Then does not change signs and is not identically either, so , contradicting the first statement. Thus there are at least distinct sign-changing roots.

## 1 Comment

1. The above idea can be employed in proving the generalized result: If is such that

then has at least distinct sign-changing roots in .