## Dodo1 2 years ago Calculus question ; photo attached

1. Dodo1

2. abb0t

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

3. Dodo1

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

4. abb0t

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

5. stamp

@abb0t that derivative is incorrect.

6. abb0t

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

7. Dodo1

how is it wrong?

8. stamp

@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

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

10. stamp

no

11. stamp

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

12. stamp

@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

oh not plus ok. I see

14. Dodo1

next step is mutiply x^2 both side?

15. Dodo1

@stamp

16. stamp

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

17. Dodo1

x=Sqrt(9/2) @stamp

18. stamp

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

19. Dodo1

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

20. Dodo1

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

21. stamp

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

23. Dodo1

Yea, I do need to prepare derivative problems. Are you going?

24. stamp

f(x) = http://www.wolframalpha.com/input/?i=9x%2B2%2Fx f'(x) = http://www.wolframalpha.com/input/?i=derivative+of+9x%2B2%2Fx f''(x) = http://www.wolframalpha.com/input/?i=second+derivative+of+9x%2B2%2Fx critical points where f'(x) is 0 = http://www.wolframalpha.com/input/?i=solve+derivative+of+9x%2B2%2Fx graph of f(x) http://www.wolframalpha.com/input/?i=plot+9x%2B2%2Fx Visually one can observe that the negative critical point is a maximum, while the positive critical point is a minimum. Analytically, you can observe the change of f''(x) to the left of the critical point and to the right of the critical point. Since f'( - c ) goes from + to - , - c is a maximum. Since f'( + c ) goes from - to +, + c is a minimum.