# Homework Help: Interesting Derivative Proof Question

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1. Feb 11, 2015

1. The problem statement, all variables and given/known data
I recently searched around SE, and found:

http://math.stackexchange.com/questions/1142546/how-to-solve-this-derivative-of-f-proof

2. Relevant equations

Below

3. The attempt at a solution

"A function given that $$f(x)=f''(x)+f'(x)g(x)$$ could be an exponential function, sine, cosine , quadratic polynomial or $$f\equiv0$$. So we can say that the function is a continuous function $\in C^2$.

The right negation is that $$f(x)\ge0 , \in (a, b)$$ and exist a point c | $$f(c)>0$$.

You have that $$f''(x_1)\ge0$$ (the function in that point is convex) so in that point you have a minima so there are two case

1. $$f(x_1)<0$$ (obviously contradiction)
2. $$f(x_1)=0$$ (it's impossible because this imply that f(x) = 0)

Analog for the other case"

But how does $$f''(x_1) > 0$$ show that $$f(x_1) < 0$$? I dont understand the contradiction for #1??

Thanks!

2. Feb 11, 2015

### PeroK

The proof is saying that if f is not identically 0, then it must attain its maximum (or minimum) at x_1 in (a, b). And f''(x_1) = f(x_1) . But, using the second derivative test, if f(x_1) > 0, then we have a local minimum and if f(x_1) < 0, we have a local maximum. Contradiction.

3. Feb 11, 2015

### epenguin

Finding this strange. This should generate an infinite number of zeroes in the interval?

4. Feb 12, 2015

But how is that a contradiction and to what?

5. Feb 12, 2015

### PeroK

It's basic calculus. If f(x_1) > 0, how can that be a local minimum?

6. Feb 12, 2015

$$f(x) = x^2 + 1$$?

It has a minimum f(x_1) > 0?

7. Feb 12, 2015

### PeroK

We're not talking here about an arbitrary function. We're talking about one that meets the key criteria stated in the problem. The first of these is that f(x) = 0 at two distinct points. The second criterion is that f(x) = f''(x) whenever f'(x) = 0.

If such an f has a minimum it must be < 0. And f'' must be < 0 at that point. But, f'' < 0 implies it's a local maximum. That's the contradiction.

8. Feb 12, 2015

Ok, this is starting to make sense.

The only part I do not understand is.

"If such an f has a minimum it must be < 0"

Why does this have to be true? Thanks!

9. Feb 12, 2015

### PeroK

To be more precise, the proof was:

If f(x) < 0 for some x in (a, b), then f attains its absolute minimum on (a, b).

And, if f(x) > 0 for some x in (a, b), then f attains its absolute maximum on (a, b).

10. Feb 12, 2015

???

How if f(x) < 0 it is a contradiction to have f"(x) > 0 ??

11. Feb 12, 2015

### PeroK

It isn't! Lots of functions have that property. But NOT the one in the question which meets certain criteria.

Do you not understand that a proof may apply to a function with certain properties, but not to all functions?

12. Feb 12, 2015

I do understand it.

But can you show that for our function it is a contradiction that,

if f(x) < 0 tthen the contradiction is to have f''(x) > 0??

13. Feb 12, 2015

### PeroK

That's not correct either. We have f(x) < 0, f'(x) = 0 and f''(x) = f(x) < 0.

14. Feb 12, 2015

Yes but how will it be a contradiction?

15. Feb 13, 2015

### epenguin

Is the misunderstanding here that Amad is trying to prove a statement true by contradiction, while PeroK has proved it false by contradiction?
Perhaps the problem has been wrongly copied or misunderstood, for which there are other indications.

Also I saw earlier without even looking at the posts that if a function is such that between any two zeroes it has another zero, then it will have an infinite number of zeroes. Which is not impossible, but it would be a funny function to have an excercise on at this level.