## Dodo1 Group Title Calculus question ; photo attached one year ago one year ago

1. Dodo1 Group Title

2. abb0t Group Title

Start by taking the derivative of your function. That will be your first step. Can you do that?

3. Dodo1 Group Title

9+(2/x^2)=0?

4. abb0t Group Title

Great. Now that you have that, solve for "x".

5. stamp Group Title

@abb0t that derivative is incorrect.

6. abb0t Group Title

Hmm, is it now? Well, @Dodo1 you need to try again then.

7. Dodo1 Group Title

how is it wrong?

8. stamp Group Title

@abb0t Good luck evaluation that term = 0 without using imaginary numbers. @Dodo1 Forgot his - sign when taking the derivative using the power rule

9. Dodo1 Group Title

9+(2x^(-2))=0?

10. stamp Group Title

no

11. stamp Group Title

$9x+2x^{-1}$$9+(-1)2x^{-1-1}$$9-2x^{-2}$$9-2/x^2$

12. stamp Group Title

@Dodo1 How did you plan on solving 9 + 2/x^2 = 0? It is not possible using real numbers because any real number squared is positive, this was the first clue that something was wrong with the derivative

13. Dodo1 Group Title

oh not plus ok. I see

14. Dodo1 Group Title

next step is mutiply x^2 both side?

15. Dodo1 Group Title

@stamp

16. stamp Group Title

@Dodo1 It is algebra I at this point. Solve for x.

17. Dodo1 Group Title

x=Sqrt(9/2) @stamp

18. stamp Group Title

$9-2/x^2=0$$9=2/x^2$$x^2=2/9$$x=+/-(\sqrt2/3)$

19. Dodo1 Group Title

Ops sorry i did 9X^2+2=0

20. Dodo1 Group Title

whats the next step? max at _sart(2)/3? @stamp

21. stamp Group Title

You do not need to tag me in every response, Dodo, I am here. These values, x = +/- c, are critical points. They are max or mins, we do not know until we consider each point with respect to the second derivative. You have been familiarized with the second derivative test in lecture, now we are using it to see which of these critical points are max, mins, etc.

22. stamp Group Title