Suppose that f(x) = frac{1}{x}. Show that there is no real number c in [-1,2] such that frac{f'(c) = f(2)-f(-1)}{2-(-1)}.

Question

anonymous · Answer

$$f(x) = \frac{1}{x}$$
Show that there is no real number c in [-1,2] such that $$f'(c) = \frac{f(2)-f(-1)}{2-(-1)}$$
Why does this not contradict the mean value theorem? f(x) is not continuous on [-2,2].

anonymous · Answer

I don't really understand how to approach this problem.

anonymous · Answer

work strictly with numbers on the right
the derivative on the left is $-\frac{1}{x^2}$

anonymous · Answer

I solved for c and got $$c = \pm \sqrt{2}$$
I'm not sure how this shows there are no real roots in [-1,2] though, as it seems there is at least one! Right?

anonymous · Answer

i hope it is clear what i mean
the right hand side of the equal sign is a number 
what did you get?

anonymous · Answer

Ah never mind, I screwed up the arithmetic. I think it ends up as 
$$x^{2} = -2$$
so there are no real roots, only complex.

anonymous · Answer

yes, and so no solution

anonymous · Answer

Thanks :)

anonymous · Answer

and your answer above is correct, it is not continuous on that interval