Assume that f is non-negative on (0,1)

[mansi]

Hello...I need help and I know this is a very simple problem...I don't know why I'm getting stuck( Maybe because it's past midnight here )

Assume that f is non-negative on (0,1) and the third derivative of f exists on (0,1).If f(x)=0 for at least 2 values of x in (0,1), show that there exists "c" in (0,1) such that the third derivative of f at c is 0.

this is what I've done...let a,b be the points where f vanishes. that is f(a)=f(b)=0
so we can use Rolle's theorem and get a point p such that f^(p)=0.

What next??

 

[matt grime]

draw a picture: hint it is non-negative and touches the x-axis twice (necessarily at local minima)

 

[tacman]

First of all, don't worry about getting stuck on a problem. It's completely normal and happens to everyone, especially when it's late at night. Take a deep breath and approach the problem with a clear mind.

Now, let's continue with your solution. Since we have established that there exists a point p such that f^(p)=0, we can use Rolle's theorem again on the interval (a,p) and (p,b). This will give us two more points q and r such that f^(q)=f^(r)=0.

Now, let's consider the interval (q,r). Since f is non-negative on (0,1), f^(3) will also be non-negative on (0,1). This means that if we take the average of q and r, let's call it c, then f^(3)(c) will be greater than or equal to 0.

But we also know that f^(3)(q)=f^(3)(r)=0, so f^(3) must have a local maximum or minimum at c. Since f^(3) is continuous on (0,1) and has a local maximum or minimum at c, it must also have a critical point at c. This means that f^(3)(c)=0.

Therefore, we have shown that there exists a point c in (0,1) such that the third derivative of f at c is 0. I hope this helps! Just remember to take your time and approach the problem step by step. Good luck!