## chochko Group Title find the critical numbers f(x)=-x^4+4x^3+5 one year ago one year ago

1. ParthKohli Group Title

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

2. ParthKohli Group Title

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

3. chochko Group Title

-4x^3+12x^2

4. chochko Group Title

ANd then?

5. ParthKohli Group Title

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

6. ParthKohli Group Title

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

7. chochko Group Title

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

8. ParthKohli Group Title

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

9. chochko Group Title

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

10. Yahoo! Group Title

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

11. ParthKohli Group Title

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

12. ParthKohli Group Title

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

13. chochko Group Title

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

14. ParthKohli Group Title

Sure.

15. Mimi_x3 Group Title

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

16. ParthKohli Group Title

You guys must be in the same school.

17. ParthKohli Group Title

lol

18. chochko Group Title

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 Group Title

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 Group Title

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

21. chochko Group Title

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 Group Title

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

23. chochko Group Title

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

24. Mimi_x3 Group Title

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

25. chochko Group Title

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

26. Mimi_x3 Group Title

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 Group Title

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

28. Mimi_x3 Group Title

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 Group Title

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

30. Mimi_x3 Group Title

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

31. Mimi_x3 Group Title

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

32. chochko Group Title

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 Group Title

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

34. chochko Group Title

ok thanks a lot

35. chochko Group Title

@ParthKohli are you able to help me finish this one

36. Mimi_x3 Group Title

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 Group Title

Sry, back.

38. ParthKohli Group Title

Didn't notice the tag right there. :P

39. chochko Group Title

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

40. ParthKohli Group Title

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

41. ParthKohli Group Title

What is the question again?

42. chochko Group Title

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 Group Title

That's right.

44. ParthKohli Group Title

So it starts decreasing after 3.

45. chochko Group Title

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

46. ParthKohli Group Title

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 Group Title

It's decreasing at f(-5).

48. ParthKohli Group Title

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

49. chochko Group Title

huh? whos that?

50. ParthKohli Group Title
51. ParthKohli Group Title

Enter your function and see its graph.

52. chochko Group Title

ok can you help me find the possible inflection points

53. chochko Group Title

@ParthKohli can you help just a lil bit left

54. ParthKohli Group Title

Sorry, back.

55. ParthKohli Group Title

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