Spring 2004 #5

Problem Statement: Let f:\mathbb{R}\rightarrow\mathbb{R} be differentiable. Let a,b\in \mathbb{R} such that a<b and f'(a)\neq f'(b). If N is between f'(a) and f'(b) then there exists M\in (a,b) such that f'(M)=N.

Proof: Wlog assume f'(a)<f'(b). Fix N\in (f'(a),f'(b)). Define g(x)=Nx-f(x). Then g'(x)=N-f'(x). Note that g'(a)=N-f'(a)>0 and g'(b)=N-f'(b)<0 since f'(a)<N<f'(b).

g'(a)>0 and so for x sufficiently close to a, g(x)>g(a) for x>a. Similarly, g'(b)<0 and so for x sufficiently close to x, g(x)>g(b) for x<b

Since g is continuous on [a,b], which is a compact set, we know that g attains both it’s max and min on [a,b]. We know from above that g  does not attain it’s max when x=a or x=b. So, let M\in (a,b) such that g(M)\geq |g(x)| for every x\in [a,b]. Then g(M) is the maximum, thus, g'(M)=0. Thus, g'(M)=N-f'(M)=0 and so it follows that N=f'(M)


Reflection: This one was not straightforward at all. I think that after doing this problem I would see the trick, but right away I didn’t see it. When I tried to attack it directly I didn’t make any real progress on the proof. So, maybe this is a good technique for me to have up my sleeve; when trying to prove something about the derivative create a function you know information about.

This entry was posted in Analysis, Compact, Continuity, Differentiable, Math. Bookmark the permalink.

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