Ok. Next time be more precise then . Take a look at Mean value theorem - Wikipedia, the free encyclopedia. You must prove this theorem... called the mean value theorem. (or something very close to it).
Yes. The mean value theorem states that there exist a in such as . In your example, and are defined and continuous on , you must find a subinterval of in which the mean value theorem can be applied.is there any connection between mean value theorem and the thing we asked to prove ?
The intermediate value theorem for derivatives states that if is a differenciable function then has the intermediate value property on , an open interval. Meaning if with and lies (strictly) between and there exists so that .
Without lose of generality we will assume that . Now define the function as . Since is continous on it assumes a minimum value. We will now show that does not take a minimum value at . Note that . But since it follows that for sufficiently close to . Thus, there exists so that , which is means , in particular is not a minimum for . Using a similar argument we can show is not a minimum for . Thus, has a minimum on and so there is a point so that .