1. Intermediate value theorem

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.

2. weird.. it must have been easier if it was $f'(c) - f'(d)$..
hahaha.. i am looking forward for the solution..

3. Further workings.

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.

4. Originally Posted by spammanon
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.
I haven't had a really hard look, but the Mean Value Theorem might come into it.

5. 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 $\mathrm{f}(2)\ne\frac{\mathrm{f}(4)+\mathrm{f}(0)} {2}$.

Define $\mathrm{g}:[0,2]\to\mathbb{R}$ by $\mathrm{g}(x)=\mathrm{f}(x+2)-\mathrm{f}(x)-\frac{\mathrm{f}(4)-\mathrm{f}(0)}{2}$.

Thus:

${g}(0)=\mathrm{f}(2)-\frac{\mathrm{f}(4)}{2}-\frac{\mathrm{f}(0)}{2}$

${g}(2)=-\mathrm{f}(2)+\frac{\mathrm{f}(4)}{2}+\frac{\mathr m{f}(0)}{2}$

Hence if $\mathrm{f}(2)\ne\frac{\mathrm{f}(4)+\mathrm{f}(0)} {2}$, ${g}(0)$ and ${g}(2)$ are nonzero and opposite in sign; hence by the IVT, $\exists\,d\in[0,2]$ such that ${g}(d)=0$.

So it remains to prove the result for the case $\mathrm{f}(2)=\frac{\mathrm{f}(4)+\mathrm{f}(0)}{2 }$,

6. Looks like you have almost got it.

Thanks for your work so far! I will keep going and see if I can figure out the rest from what you posted. I would appreciate it if you found the complete proof too!