# First and Second Derivatives

1. Apr 21, 2012

### Bipolarity

Can this problem be solved? I made up the problem myself so I am not sure a solution exists.

It is known that:
$$h(0) = 0$$
$$h'(0) < 0$$
$$h'' > 0$$

Prove that there exists a value $c > 0$ such that $h(c)=0$

It makes sense visually. I have tried applying the MVDT/IVT in various ways, but its not hitting me. Perhaps someone can start me off?

BiP

2. Apr 21, 2012

### DrewD

That's not a counter example since its derivatives are equal to $0$ at $x=0$

3. Apr 21, 2012

### Useful nucleus

The derivatives of h(x)=-$\sqrt{x}$ are not definied at x=0.
EDIT: I have just noticed that the question requires h/(0)<0. So my example is not relevant and I deleted my post.

Last edited: Apr 21, 2012
4. Apr 21, 2012

### DrewD

yes, undefined, sorry.

5. Apr 21, 2012

### Useful nucleus

Consider h(x) = -ln(1+x) defined on (-1,∞). In this case h(0)=0, h/(0)=-1<0, h//(x)= (1+x)-2 >0 when x$\in$(-1,∞). However, h(x)≠0 when x>0.

Last edited: Apr 21, 2012
6. Apr 21, 2012

### chiro

Hey BiPolarity.

I think the answer to your question is an emphatic yes based on the following argument:

At t = 0, h(t) = 0. The derivative is negative which means the function is decrease after t = 0, however since h'' > 0, this implies that the derivative will always increase which means that at some point the derivative will become positive after a turning point which will be a minimum in which the function will accelerate positively for the rest of the values of t after this turning point.

So what you can do is firstly show that there is a turning point at some value t = x where x is finite, and then show that if h is continuous and analytic then since h is always increasing from h(x), then it must cross the x-axis and hence there exists a root at some finite value of t = x + y for some finite y where x is the solution for the turning point.

So you could show in your proof that a) a turning point exists for the finite value of t = x, and that b) the function is always increasing after t = x, and that b) there exists a root for some finite value of t = x + y given that h is continuous and always is monotonically increasing from t = x onwards.

7. Apr 21, 2012

### Bipolarity

Okay, I think I see how this would work out. Two questions:
1) What does "analytic" mean?
2) Can I prove that if $f' > 0$ and $f(0) < 0$ then for some value $c > 0$ it must be true that $f(c) = 0$. If I can solve this then I have essentially solved the problem. But I think solving this problem requires axioms that I may not have studied yet. I have not taken any analysis courses yet.

BiP

8. Apr 21, 2012

### Useful nucleus

This need not be true. You assume that the first derivative increases without bound and can reach zero. The counter example I gave (h(x)= -ln(1+x)) show that the frist derivative will never reach zero.

9. Apr 21, 2012

### chiro

For 1) Analytic means that you have a smooth function in that the derivative exists for the appropriate domain of the function at all points and that the derivative function is also continuous. If it's not continuous it's not analytic (in this definition).

For 2) yes, this is pretty much what I was getting at. In fact there are probably a few different ways you could do this and one way would be to use the fundamental theorem of calculus in showing that after the turning point, the integral of f'(x)dx will always be positive from our turning point to some later value. If you can show that there is a solution to this then you're done.

Now we know that at the turning point h(t = x) = u < 0 which is finite. If you can show that integral f'(x)dx (integral of derivative) from t = x to t = x + y to be -u then you're done.

10. Apr 21, 2012

### lugita15

Not necessarily. Just because the derivative is increasing does not mean it will eventually become positive. In general, just because a function is increasing does not mean it will eventually attain any value. Consider a function with a horizontal asymptote: it can be increasing forever, but it may never go above the asymptote.
Even if h is increasing after t=x, it may have a horizontal asymptote below the x-axis, and so it may never have a root.

11. Apr 21, 2012

### chiro

If you had a horizontal asymptote, then the second derivative should be decreasing not increasing. Remember that f'' measures the rate of change of the derivative so if this were the case then f'' would be < 0 but it's not because this condition says f'' > 0. So this won't happen if f'' > 0 is forced.

12. Apr 21, 2012

### lugita15

Consider the function h(x)=e^(-x) - 1. It satisfies h(0)=0, h'(0)=-1<0, and h''(x)=e^(-x)>0 for all x. Yet there is no value c>0 such that h(c)=0.

13. Apr 21, 2012

### lugita15

But just because f''>0 doesn't mean f'' is increasing. In order for the second derivative to be increasing, the third derivative must be positive, which need not be true. So you can have a situation where the first derivative is initially negative and the second derivative is always positive, but the second derivative is decreasing, so that the first derivative keeps increasing but it increases more and more slowly, and thus the first derivative can have a horizontal asymptote.

14. Apr 21, 2012

### chiro

Yeah you're right, thank you for that.

For the OP if you want to prove your statement, you will probably need to add that your function has a turning point of some sort: in other words, there exists a situation where h'(x) = 0 for some value t = x. If this is not the case, then it won't work.

So add to your condition that if h'(x) = 0 for some finite x > 0, then and only then will this work.

15. Apr 21, 2012

### lugita15

That's true, because if the first derivative reaches zero, then the worst thing that can happen is for the first derivative to have a horizontal asymptote above zero. But if the first derivative approaches a positive constant, then the original function will approach a line with positive slope, and such a line must cross the x-axis.