Let have a third derivative that exists at all points. Show that there must exist at least one point for which
Hmm
Restrict the function to , where are increasing in that interval and for . Since is continuous, by MVT, there exists such that . Then . Since is continuous, by MVT, there exists such that . Then . Since is continuous, by MVT, there exists such that . Then . Since are increasing in that interval, therefore and . These imply that . Hence the conclusion holds.
This could be one possible case out of so many different cases. Anyway, is this particular proof correct?
Please advise. Thank you.
Here are some vague thoughts, which you may be able to convert into a proof.
First, if any one of the four functions vanishes at some point , then and so the result is proved. It follows from the intermediate value theorem that each of the (continuous) functions always has the same sign. By Darboux's theorem the same is true for .
Replacing by if necessary, you can assume that is always positive
Now suppose that is always negative. By considering what happens as , show that must be positive. The same argument, repeated, will show that is negative. Therefore the product must be positive.
By considering what happens as , use the same ideas to show that if is positive then all four functions must be positive, hence so is their product.