## chochko 3 years ago find the critical numbers f(x)=-x^4+4x^3+5

1. ParthKohli

As they always say: find $$f'(x)$$ and equate that to $$0$$.

2. ParthKohli

@chochko What would $$f'(x)$$ be here?

3. chochko

-4x^3+12x^2

4. chochko

ANd then?

5. ParthKohli

Well that's correct - now equate it to zero.

6. ParthKohli

Solve this equation:$\rm 4x^3 + 12x^2 = 0$

7. chochko

-4x^3+12x^2=0

8. ParthKohli

Yeah, I meant -4x^3 + 12x^2 = 0

9. chochko

sry my computer is really slow. Would you factor out a 4?

10. Yahoo!

Yu.. 4x^2 ( x-3) = 0 4x^2 = 0 x = 0 x-3 = 0 x = 3

11. ParthKohli

Mine too, and you can factor 4x^2 out.

12. ParthKohli

Yep, and what Yahoo! said are the critical points. ;)

13. chochko

thank you but im not done . that is just one part..can you help yahoo or ParthKohli?

14. ParthKohli

Sure.

15. Mimi_x3

the next part is to differentiate it again :P to determine its nature

16. ParthKohli

You guys must be in the same school.

17. ParthKohli

lol

18. chochko

using the second derivitive test which is -12x^2+24x=0 set to 0 to determine all relative extrema, indicating the x and Y values and wheteher is a max or a min. Help please

19. Mimi_x3

lol @ParthKohli: the next step is common sense. @chochko: with the differentiataed function sub the $$x$$ that you found in the first step if its a negative then its concave down => max. if its concave up then => min

20. ParthKohli

Sorry, was on another question, and yeah - that's the way I was taught too.

21. chochko

so in -12x^2+24x Sub the x for the critical numbers and the result is the relative extrema? is that the Max and Min as well?

22. Mimi_x3

max or min concave up => min concave down => max

23. chochko

so which part is the relative extrema? i really suck at math btw?

24. Mimi_x3

well i googled; relative extrema and max/ min extrema is the same thing

25. chochko

f''(0)=0 min f''(3)=-36 Max Is this correct?

26. Mimi_x3

i dont know; i dont have a calculator with me but if its you have to check if it has a horizontal point of inflexion

27. chochko

Ok, so the intervals are the critical points right? where are they increasing and decreasing?

28. Mimi_x3

critical points are the stationary points on the curve since $$f(x)=0$$ you differentiate it twice to determine if its increasing or decreasing increasing $$f''(x) >0$$

29. Mimi_x3

i think i made a mistake; increasing $$f'(x)>0$$

30. Mimi_x3

when the function is decreasing $$f'(x)<0$$

31. Mimi_x3

i mean you differetiate it once; to see if its increasing or decreasing sorry; im sick

32. chochko

so using the f'(x)=0 and plug in 0 and 3 for x to find where f(x) is increasing or decreasing?

33. Mimi_x3

oh no no im getting tired @ParthKohli: may like to explain :)

34. chochko

ok thanks a lot

35. chochko

@ParthKohli are you able to help me finish this one

36. Mimi_x3

I think I should explain; since Parth is hiding :P $$f'(x) = 0$$ you are finding the $$x$$ values that is the STATIONARY point. $$f'(x) >0$$ is the $$x$$ values were the function is INCREASING $$f'(x)<0$$ is where the function is DECREASING. The second derivative is the deirivative of the derivative. The curve $$y=f(x)$$ is concave up when $$f''(x)>0$$ and concve down when $$f''(x)<0$$

37. ParthKohli

Sry, back.

38. ParthKohli

Didn't notice the tag right there. :P

39. chochko

Increasing on ( -inf, 0) (0,3) then decreasing (3, inf) idk if its correct?

40. ParthKohli

Well your $$\rm f'(x)$$ function is $$\rm -4x^2 +12x$$ and the function is stationary at x = 3,0.

41. ParthKohli

What is the question again?

42. chochko

State the intervas for which f(x) is increasing or decreasing So i tested the values -1, 1, 4 And got F'(-1)=16 inc f'(2)=8 inc f'(4)=-64 dec. idk if its correct?

43. ParthKohli

That's right.

44. ParthKohli

So it starts decreasing after 3.

45. chochko

(-inf, 0) (0,3) increasing then (3, inf) decreasing Are the intervals right?

46. ParthKohli

Let's test f'(-5). f'(x) = -4x^2 + 12x f'(-5) = -4(-5)^2 + 12(-5) = -4(25) - 60 = -100 - 60 = -160

47. ParthKohli

It's decreasing at f(-5).

48. ParthKohli

The best way is to ask Wolfram for the graph. :)

49. chochko

huh? whos that?

50. ParthKohli
51. ParthKohli

Enter your function and see its graph.

52. chochko

ok can you help me find the possible inflection points

53. chochko

@ParthKohli can you help just a lil bit left

54. ParthKohli

Sorry, back.

55. ParthKohli

Damn... I've disabled the tag notifs.