# Interesting Derivative Proof Question

## Homework Statement

I recently searched around SE, and found:

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

Below

## 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!

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PeroK
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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.

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

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.
But how is that a contradiction and to what?

PeroK
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But how is that a contradiction and to what?
It's basic calculus. If f(x_1) > 0, how can that be a local minimum?

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

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

It has a minimum f(x_1) > 0?

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

It has a minimum f(x_1) > 0?
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.

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.
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!

PeroK
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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!
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).

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).

???

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

PeroK
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???

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

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?

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??

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

Yes but how will it be a contradiction?
hat's not correct either. We have f(x) < 0, f'(x) = 0 and f''(x) = f(x) < 0.

epenguin
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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.