Thread: Another derivative proof

Could someone please check this proof for me?


Question: Suppose f is a differentiable function. Prove that if f(0) = 0 and then f(x) = 0 for x in (0,1).

Proof:


Assume there exists a c in (0,1) such that f(c) > 0. Then by the MVT there exists a c* in (0,c) such that . We know that . So since c is in (0,1).

It follows that f(c) < f(c*) whenever c* < c. But this means that the function is decreasing. However, f(0) = 0 and from our assumption that f(c) > 0, this must mean that f must be increasing towards f(c), which is a contradiction. The proof for assuming f(c) < 0 is similar.

I think the proof is a bit shaky in the last paragraph. How is it?

Originally Posted by JG89

Could someone please check this proof for me?


Question: Suppose f is a differentiable function. Prove that if f(0) = 0 and then f(x) = 0 for x in (0,1).

Proof:


Assume there exists a c in (0,1) such that f(c) > 0. Then by the MVT there exists a c* in (0,c) such that . We know that . So since c is in (0,1).

It follows that f(c) < f(c*) whenever c* < c. But this means that the function is decreasing. However, f(0) = 0 and from our assumption that f(c) > 0, this must mean that f must be increasing towards f(c), which is a contradiction. The proof for assuming f(c) < 0 is similar.

I think the proof is a bit shaky in the last paragraph. How is it?

There is an perfect solution!
Assume

Let ,we get

Thanks ynj!

How did you know to write down those inequalities though? Was it by trial-and-error?

Originally Posted by JG89

Thanks ynj!

How did you know to write down those inequalities though? Was it by trial-and-error?

Because I have seen this problem in my book before.