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anonymous

  • 5 years ago

how do i find the critical numbers for cube root of X+2 . I took the derivative of this and got (1)/[3(x+2)^(-2/3)]. is the next step setting this expression equal to 0?

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  1. anonymous
    • 5 years ago
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    yes

  2. anonymous
    • 5 years ago
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    but Im having problem solving for 0 at this point.

  3. anonymous
    • 5 years ago
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    one sec

  4. anonymous
    • 5 years ago
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    and consider the x that will make the equation undefined as a critical number

  5. anonymous
    • 5 years ago
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    make the derivative i mean

  6. anonymous
    • 5 years ago
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    so I have to substitute a value for x that will make this expression ---> (1)/[3(x+2)^(-2/3)]. equal 0?

  7. anonymous
    • 5 years ago
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    your derivative should be 3x^2+12x+12

  8. anonymous
    • 5 years ago
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    if I substitute -2 for x, the entire denominator equals 0. so would -2 b my critical number?

  9. anonymous
    • 5 years ago
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    yes you cant find a number that can make your derivative equal to 0 so -2 is the only critical number

  10. anonymous
    • 5 years ago
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    ohh okay!!! thanks a lot! :)

  11. anonymous
    • 5 years ago
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    @pineda isnt the derivative wrong?

  12. anonymous
    • 5 years ago
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    I think its right. I double checked it on my calculator

  13. anonymous
    • 5 years ago
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    it looks fine for me its \[1/3\sqrt[3]{x+2^{2}}\]

  14. anonymous
    • 5 years ago
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    is the equation \[(x+2)^{3}\]

  15. anonymous
    • 5 years ago
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    pineda100 is right

  16. anonymous
    • 5 years ago
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    its \[\sqrt[3]{x+2}\]

  17. anonymous
    • 5 years ago
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    yeah

  18. anonymous
    • 5 years ago
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    oh oops read it wrong

  19. anonymous
    • 5 years ago
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    ^_^

  20. anonymous
    • 5 years ago
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    original equation is \[\sqrt[3]{ }x+2\]. sorry i coudlnt get the square root sign to appear completely

  21. anonymous
    • 5 years ago
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    ya then pineda is correct. my bad.

  22. anonymous
    • 5 years ago
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    okay thanks a lot for your help pineda and everyone else who contritbuted! much appreciated! :)

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