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mathcalculus

  • 3 years ago

HELPPPPPPP!!!! locate all critical points (both types) of h(x)= radical 8-6*x^2 The critical point(s):

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  1. mathcalculus
    • 3 years ago
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    I FOUND THE X WHICH IS 0. THEN I TRIED TO FIND Y BY PLUGGING IN ZERO BUT IT IS INCORRECT... :(

  2. mathcalculus
    • 3 years ago
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    i know the steps to this problem. which is find the derivative and i did...

  3. mathcalculus
    • 3 years ago
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    then = to zero. and i did that too. and I goy x but then im not sure how i got the answer wrong....

  4. zepdrix
    • 3 years ago
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    It's hard to read the function when it's not formatted. Does this look accurate?\[\large h(x)=\sqrt{8-6x^2}\]Everything under the square root like that?

  5. mathcalculus
    • 3 years ago
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    yes

  6. zepdrix
    • 3 years ago
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    Hmm so we get something like this, yes? \[\large h'(x)=\frac{-6x}{\sqrt{8-6x^2}}\] So you determined that \(\large x=0\) is a critical point. There is also another. A critical point of the function also exists where \(\large f'(x)\) is undefined. (Assuming that value is included in the domain of \(\large f\).

  7. mathcalculus
    • 3 years ago
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    okay.

  8. mathcalculus
    • 3 years ago
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    yeah I determined one critical point, but what about y?

  9. zepdrix
    • 3 years ago
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    I'm saying that there is another critical point,\[\large \sqrt{8-6x^2}\]When this thing equals zero, we're dividing by zero, so f'(x) is undefined, DNE. \[\large 0=\sqrt{8-6x^2}\] Solve for x to find your other critical point(s).

  10. zepdrix
    • 3 years ago
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    As for your y values, if you want to, you can plug x=0 into your original function, to find the corresponding y value. That will allow you to write it as an ordered pair if you want.

  11. mathcalculus
    • 3 years ago
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    yes i found 2 radical 2

  12. mathcalculus
    • 3 years ago
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    and i wrote the answer like this... (0,2 radical 2)

  13. zepdrix
    • 3 years ago
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    Hmm I think you end up with \(\large x=\pm \dfrac{2}{\sqrt3}\)

  14. mathcalculus
    • 3 years ago
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    (0,[2\sqrt{2}\])

  15. mathcalculus
    • 3 years ago
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    how?

  16. mathcalculus
    • 3 years ago
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    so x is not 0?

  17. zepdrix
    • 3 years ago
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    \[\large 0=8-6x^2\]\[\large 8=6x^2\]\[\large \frac{8}{6}=x^2 \qquad \rightarrow \qquad \frac{4}{3}=x^2\]\[\large x=\pm \sqrt{\frac{4}{3}}\qquad \rightarrow \qquad x=\pm \frac{2}{\sqrt3}\]

  18. mathcalculus
    • 3 years ago
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    to find the critical points.. dont we find the derivative first?

  19. zepdrix
    • 3 years ago
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    Yes, this was our derivative,\[\large h'(x)=\frac{-6x}{\sqrt{8-6x^2}}\]We set the numerator equal to zero to find one critical point. We need to also set the denominator equal to zero to find the other(s). Which is what I was showing.

  20. mathcalculus
    • 3 years ago
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    okay... then

  21. mathcalculus
    • 3 years ago
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    we solve the denominator = to 0 but what about the radical we ignore it?

  22. mathcalculus
    • 3 years ago
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    have you used webwork?

  23. mathcalculus
    • 3 years ago
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    im not sure how to plug in the answer

  24. mathcalculus
    • 3 years ago
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    (0,

  25. zepdrix
    • 3 years ago
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    The square root? I squared both sides. 0^2 = 0

  26. mathcalculus
    • 3 years ago
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    no the answer; how do we write it if there a positive and a negative 2 radical 3.

  27. zepdrix
    • 3 years ago
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    I've used webassign, not webworks. Hmm Is there a panel on the left somewhere where you can use tools like sqrt?

  28. mathcalculus
    • 3 years ago
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    the critical points are:

  29. mathcalculus
    • 3 years ago
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    ive tried every way but it's "Wrong"

  30. mathcalculus
    • 3 years ago
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    yes sqrt( )

  31. zepdrix
    • 3 years ago
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    Write it as three separate critical points: (0, sqrt(8)) (2/sqrt(3),0) (-2/sqrt(3),0) Does that work maybe? D:

  32. mathcalculus
    • 3 years ago
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    Operands of '*' are not of compatible types

  33. mathcalculus
    • 3 years ago
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    because its between of (0, sqrt(8)) (2/sqrt(3),0) (-2/sqrt(3),0)

  34. mathcalculus
    • 3 years ago
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    hold on im so confused.... lol how did you get (0,8)

  35. mathcalculus
    • 3 years ago
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    (0, sqrt8)?

  36. zepdrix
    • 3 years ago
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    We found that x=0 is a critical point. If we plug x=0 back into the original function it gives us, \(\large f(0)=\sqrt8\) right? We can write that as an ordered pair (0,sqrt(8))

  37. zepdrix
    • 3 years ago
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    "Operands of '*' are not of compatible types" I don't understand what you're saying :o

  38. mathcalculus
    • 3 years ago
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    not me, thats what webworks says..

  39. zepdrix
    • 3 years ago
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    Does it tell you to list them as ordered pairs? Usually you just list critical points as x=

  40. mathcalculus
    • 3 years ago
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    Locate all critical points ( both types ) The critical point(s) is (are) . (The answers are to be points. Use parentheses in your answer(s). If there are no critical points, enter none .)

  41. mathcalculus
    • 3 years ago
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    now i tried to put the answer like this: (0,sqrt(8))U(2/sqrt(3),0)U(-2/sqrt(3),0) but it says: Left endpoint must be less than right endpoint

  42. zepdrix
    • 3 years ago
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    Hmm I'm not sure how your assignment wants them formatted. I would guess that you just separate them with commas. (0,sqrt(8)), (2/sqrt(3),0), (-2/sqrt(3),0)

  43. mathcalculus
    • 3 years ago
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    correctttt :)

  44. mathcalculus
    • 3 years ago
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    okay, so after we find x, how do we get the + and - 2 radical 3? i mean we plug in 0 to the original but i got 2 radical 2

  45. mathcalculus
    • 3 years ago
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    i understand the 0 and sqrt 8 order pair...

  46. mathcalculus
    • 3 years ago
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    why couldnt i write 0, 2 radical 2 as a critical point?

  47. zepdrix
    • 3 years ago
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    Scroll up to see my steps. I don't want to write them out again. Set the denominator equal to zero, solve for x.

  48. mathcalculus
    • 3 years ago
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    nvm but now... how about the +&- 2 radical 3.

  49. zepdrix
    • 3 years ago
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    We have an \(\large x^2\). When we take the root of a variable, we get two solutions, the positive and negative root.

  50. mathcalculus
    • 3 years ago
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    i understand how you made the cal. but why did you ignore the radical and wrote it as 8-6x^2

  51. mathcalculus
    • 3 years ago
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    yes ,right. we have a postive and negative. i understand that but why did you ignore the radical and wrote it as 8-6x^2? when the original one had a radical .

  52. zepdrix
    • 3 years ago
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    I didn't ignore the radical.\[\large 0=\sqrt{8-6x^2}\]Step 1, square both sides.

  53. mathcalculus
    • 3 years ago
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    thank you so much! :)

  54. mathcalculus
    • 3 years ago
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    thank you thank you! ^_^

  55. mathcalculus
    • 3 years ago
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    i'll keep that the radicals need a + and - in mind.

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