weird.. it must have been easier if it was ..
hahaha.. i am looking forward for the solution..
Can anyone help me with this?
Let f be a continuous function on [0,4]. Show that there must be values c and d in [0,4] such that:
c - d = 2 and f(c) - f(d)= [f(4) - f(0)]/2
I have been struggling with this one for a while now. The best thing I can think of is to split [0,4] into two intervals and use the IVT to show that f(c) and f(d) exist, but I don't know how the fact that their difference is the average of the interval affects things. Thanks.
I think the problem is a special case of the general rule (at least it only makes sense):
Let f be a continuous function on [beg,end]. Show that there must be values c and d in [beg,end] such that:
c - d = (beg+end)/2 and f(c) - f(d)= [f(end) - f(beg)]/2
But I still don't know how to show this is true. I have checked with all kinds of functions for all kinds of intervals, and it appears true.
You can NOT use the mean-value theorem. The MVT is only applicable to differentiable functions but nothing is stated in the question about f being differentiable.
This is my partial proof, which works only if .
Define by .
Hence if , and are nonzero and opposite in sign; hence by the IVT, such that .
So it remains to prove the result for the case ,