## anonymous one year ago Find the absolute extrema of...

1. anonymous

$\sqrt[3]{x}$

2. anonymous

I already solved for the derivative and got... $\frac{ 1 }{ 3 }x^{-\frac{ 2 }{ 3 }}$

3. anonymous

I'm just not exactly sure what to do next.

4. Michele_Laino

by the Weierstrass theorem we need of a closed interval of the real line

5. anonymous

I can't say that I've learned about the Weierstrass theorem.

6. anonymous

Is that where you just find all the critical points of the closed interval and determine which is the max?

7. Michele_Laino

yes!

8. anonymous

Awesome! One question though, how would I find the critical points of my derivative since I can't really isolate x (at least from what I can see)?

9. Michele_Laino

your first derivative is always positive, which means that your function is an increasing function

10. Michele_Laino

of course at the point $$x=0$$ your first derivative is not defined

11. anonymous

So it would have an asymptote?

12. Michele_Laino

no, since the asymptotes occurs at point $$x$$ such that the function (not the first derivative) becomes an infinity

13. anonymous

Okay, I think I understand.

14. Michele_Laino

ok! :)

15. anonymous

Since my function is an increasing function, is there a way to have specific critical points? (like 1,2,3..)

16. anonymous

Or would it just be infinity?

17. Michele_Laino

as I said before, we need of a closed interval. For example if we have this interval: $$[2,3]$$ then we have to evaluate $$f(2),f(3)$$. We will conclude that $$f(2)$$ is the point of minimum, and $$f(3)$$ is the point of maximum for function $$f$$

18. anonymous

Okay. Got it. Thank you! :)

19. Michele_Laino

:)