# Relative extrema problem for Essentials of Calculus

1. Jul 12, 2010

### Nawz

1. The problem statement, all variables and given/known data

Find the relative extrema of each function, if they exist. List each extremum along with the x-value at which it occurs. Then sketch a graph of the function.

25. f(x)=1-x2/3

2. Relevant equations

3. The attempt at a solution

f prime of (x)= -2/3x-1/3

-2/3x-1/3= 0

??

2. Jul 12, 2010

### Staff: Mentor

There are no solutions to the equation f'(x) = 0. Extreme values can occur at points other than where the deriviative is zero. What are these other points?

3. Jul 12, 2010

### Nawz

Thanks for the help, I figured it out.

It was (x)=0 for f'(x) , idk why I couldn't see that. so when i put that in the original equation i got 1. So the x point was (0,1) . And then I found the remaning points on the graph and it was correct with the back of the book answer.

Thank you

If i have another questions, can i just ask here or should i make a new topic? It's the same type of questions just another problem?

Last edited: Jul 12, 2010
4. Jul 12, 2010

### Staff: Mentor

But x = 0 is not a solution of f'(x) = 0, and that's what you were looking for. Do you understand why x = 0 is an extreme point?
Please start a new thread.

5. Jul 12, 2010

### Nawz

No i don't know what you mean. I came back from class but I remember this problem still. (0,1) was the realtive maximum at least I am fairly sure. I don't need a solution, i just needed the relative extrema which is (0,1) and then sketch... thats right, isn't it?

6. Jul 13, 2010

### Staff: Mentor

Yes, (0, 1) is the maximum point. An extreme point of a function f can occur at any of the following:
1. Points at which f'(x) = 0
2. Points in the domain of f at which f' is undefined
3. Endpoints of the domain of f

The work you showed in your first post was toward the first point above, but for your function, there are no values x for which f'(x) = 0. It wasn't clear to me that you understood where to look for extreme values.

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