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chochko

  • 3 years ago

find the critical numbers f(x)=-x^4+4x^3+5

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  1. ParthKohli
    • 3 years ago
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    As they always say: find \(f'(x)\) and equate that to \(0\).

  2. ParthKohli
    • 3 years ago
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    @chochko What would \(f'(x)\) be here?

  3. chochko
    • 3 years ago
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    -4x^3+12x^2

  4. chochko
    • 3 years ago
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    ANd then?

  5. ParthKohli
    • 3 years ago
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    Well that's correct - now equate it to zero.

  6. ParthKohli
    • 3 years ago
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    Solve this equation:\[\rm 4x^3 + 12x^2 = 0\]

  7. chochko
    • 3 years ago
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    -4x^3+12x^2=0

  8. ParthKohli
    • 3 years ago
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    Yeah, I meant -4x^3 + 12x^2 = 0

  9. chochko
    • 3 years ago
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    sry my computer is really slow. Would you factor out a 4?

  10. Yahoo!
    • 3 years ago
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    Yu.. 4x^2 ( x-3) = 0 4x^2 = 0 x = 0 x-3 = 0 x = 3

  11. ParthKohli
    • 3 years ago
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    Mine too, and you can factor 4x^2 out.

  12. ParthKohli
    • 3 years ago
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    Yep, and what Yahoo! said are the critical points. ;)

  13. chochko
    • 3 years ago
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    thank you but im not done . that is just one part..can you help yahoo or ParthKohli?

  14. ParthKohli
    • 3 years ago
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    Sure.

  15. Mimi_x3
    • 3 years ago
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    the next part is to differentiate it again :P to determine its nature

  16. ParthKohli
    • 3 years ago
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    You guys must be in the same school.

  17. ParthKohli
    • 3 years ago
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    lol

  18. chochko
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    Sorry, was on another question, and yeah - that's the way I was taught too.

  21. chochko
    • 3 years ago
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    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
    • 3 years ago
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    max or min concave up => min concave down => max

  23. chochko
    • 3 years ago
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    so which part is the relative extrema? i really suck at math btw?

  24. Mimi_x3
    • 3 years ago
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    well i googled; relative extrema and max/ min extrema is the same thing

  25. chochko
    • 3 years ago
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    f''(0)=0 min f''(3)=-36 Max Is this correct?

  26. Mimi_x3
    • 3 years ago
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    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
    • 3 years ago
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    Ok, so the intervals are the critical points right? where are they increasing and decreasing?

  28. Mimi_x3
    • 3 years ago
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    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
    • 3 years ago
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    i think i made a mistake; increasing \(f'(x)>0\)

  30. Mimi_x3
    • 3 years ago
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    when the function is decreasing \(f'(x)<0\)

  31. Mimi_x3
    • 3 years ago
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    i mean you differetiate it once; to see if its increasing or decreasing sorry; im sick

  32. chochko
    • 3 years ago
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    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
    • 3 years ago
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    oh no no im getting tired @ParthKohli: may like to explain :)

  34. chochko
    • 3 years ago
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    ok thanks a lot

  35. chochko
    • 3 years ago
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    @ParthKohli are you able to help me finish this one

  36. Mimi_x3
    • 3 years ago
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    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
    • 3 years ago
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    Sry, back.

  38. ParthKohli
    • 3 years ago
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    Didn't notice the tag right there. :P

  39. chochko
    • 3 years ago
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    Increasing on ( -inf, 0) (0,3) then decreasing (3, inf) idk if its correct?

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

  41. ParthKohli
    • 3 years ago
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    What is the question again?

  42. chochko
    • 3 years ago
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    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
    • 3 years ago
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    That's right.

  44. ParthKohli
    • 3 years ago
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    So it starts decreasing after 3.

  45. chochko
    • 3 years ago
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    (-inf, 0) (0,3) increasing then (3, inf) decreasing Are the intervals right?

  46. ParthKohli
    • 3 years ago
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    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
    • 3 years ago
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    It's decreasing at f(-5).

  48. ParthKohli
    • 3 years ago
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    The best way is to ask Wolfram for the graph. :)

  49. chochko
    • 3 years ago
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    huh? whos that?

  50. ParthKohli
    • 3 years ago
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    http://wolframalpha.com

  51. ParthKohli
    • 3 years ago
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    Enter your function and see its graph.

  52. chochko
    • 3 years ago
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    ok can you help me find the possible inflection points

  53. chochko
    • 3 years ago
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    @ParthKohli can you help just a lil bit left

  54. ParthKohli
    • 3 years ago
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    Sorry, back.

  55. ParthKohli
    • 3 years ago
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    Damn... I've disabled the tag notifs.

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