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Math Help - Critical Number Problem

  1. #1
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    Critical Number Problem

    Find the critical numbers:

    h(t) = t^{3/4} - 3t^{1/4}

    h'(t) = \dfrac{3}{4} u^{-1/4} - (\dfrac{1}{4}) 3 v^{-3/4}

    h'(t) = \dfrac{3}{4} u^{-1/4} - \dfrac{3}{4} v^{-3/4}

    h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}

    h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4} = 0 Next move. How can I solve for t in this situation?
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  2. #2
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    Re: Critical Number Problem

    Multiply both sides by t^a where -a is the smallest exponent of t. (In other words, multiply both sides by t^{3/4}).
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    Re: Critical Number Problem

    Post Edited

    Find the critical numbers:

    h(t) = t^{3/4} - 3t^{1/4}

    h'(t) = \dfrac{3}{4} u^{-1/4} - (\dfrac{1}{4}) 3 v^{-3/4}

    h'(t) = \dfrac{3}{4} u^{-1/4} - \dfrac{3}{4} v^{-3/4}

    h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}

    h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4} = 0

    h'(t) = -\dfrac{3}{4} t^{-3/4} = -\dfrac{3}{4} t^{-1/4}

    h'(t) =  t^{-3/4} =  t^{-1/4} = t^{-4/4} = t^{-1}
    Last edited by Jason76; October 17th 2013 at 12:32 PM.
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  4. #4
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    Re: Critical Number Problem

    Quote Originally Posted by Jason76 View Post
    Post Edited

    Find the critical numbers:

    h(t) = t^{3/4} - 3t^{1/4}

    h'(t) = \dfrac{3}{4} u^{-1/4} - (\dfrac{1}{4}) 3 v^{-3/4}

    h'(t) = \dfrac{3}{4} u^{-1/4} - \dfrac{3}{4} v^{-3/4}

    h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}

    h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4} = 0

    h'(t) = -\dfrac{3}{4} t^{-3/4} = -\dfrac{3}{4} t^{-1/4}

    h'(t) =  t^{-3/4} =  t^{-1/4} = t^{-4/4} = t^{-1}
    Huh? Why would you think t^{-3/4} = t^{-4/4}?
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  5. #5
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    Re: Critical Number Problem

    Find the critical numbers:

    h(t) = t^{3/4} - 3t^{1/4}

    h'(t) = \dfrac{3}{4} u^{-1/4} - (\dfrac{1}{4}) 3 v^{-3/4}

    h'(t) = \dfrac{3}{4} u^{-1/4} - \dfrac{3}{4} v^{-3/4}

    h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}

    h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4} = 0

    h'(t) = -\dfrac{3}{4} t^{-3/4} = -\dfrac{3}{4} t^{-1/4}

    h'(t) = t^{-3/4} = t^{-1/4}

    h'(t) = \dfrac{ t^{-1/4}}{t^{-3/4}} = 0

    h'(t) = t^{1/2} = 0 ??
    Last edited by Jason76; October 17th 2013 at 06:23 PM.
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  6. #6
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    Re: Critical Number Problem

    Why do you keep writing h'(t)? When you go from the line that reads
    h'(t) = \dfrac{3}{4}t^{-1/4}-\dfrac{3}{4}t^{-3/4} = 0
    to the line that reads
    h'(t) = -\dfrac{3}{4}t^{-3/4} = -\dfrac{3}{4}t^{-1/4}, you no longer have h'(t). You subtracted \dfrac{3}{4}t^{-1/4} from every term, so you should have h'(t) - \dfrac{3}{4}t^{-1/4} = -\dfrac{3}{4}t^{-3/4} = -\dfrac{3}{4}t^{-1/4}

    So just drop h'(t) = from every line. It is not true after line 5.

    Next, why would you think that \dfrac{t^{-3/4}}{t^{-3/4}} = 0?
    Last edited by SlipEternal; October 17th 2013 at 06:40 PM.
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  7. #7
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    Re: Critical Number Problem

    Find the critical numbers:

    h(t) = t^{3/4} - 3t^{1/4}

    h'(t) = \dfrac{3}{4} u^{-1/4} - (\dfrac{1}{4}) 3 v^{-3/4}

    h'(t) = \dfrac{3}{4} u^{-1/4} - \dfrac{3}{4} v^{-3/4}

    h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}

    \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4} = 0

     -\dfrac{3}{4} t^{-3/4} = -\dfrac{3}{4} t^{-1/4}

     t^{-3/4} = t^{-1/4}

     \dfrac{ t^{-1/4}}{t^{-3/4}} = 0

     t^{1/2} = 0
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  8. #8
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    Re: Critical Number Problem

    Also, from now on, it might be easier to help you if you number each line. I am "quoting" you, but I will make changes to your quote to add line numbers and make each line correct.

    Quote Originally Posted by Jason76 View Post
    Find the critical numbers:

    1. h(t) = t^{3/4} - 3t^{1/4}

    2. h'(t) = \dfrac{3}{4} u^{-1/4} - (\dfrac{1}{4}) 3 v^{-3/4}

    3. h'(t) = \dfrac{3}{4} u^{-1/4} - \dfrac{3}{4} v^{-3/4}

    4. h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}

    5. h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}=0

    6. -\dfrac{3}{4} t^{-3/4} = -\dfrac{3}{4} t^{-1/4}

    7. t^{-3/4} = t^{-1/4}

    8. \dfrac{ t^{-1/4}}{t^{-3/4}} = 0 <-- This line is wrong

    9. t^{1/2} = ???
    Now, the question is, how did you get from line 7 to line 8?
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  9. #9
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    Re: Critical Number Problem

    Quote Originally Posted by Jason76 View Post
    Post Edited

    Find the critical numbers:

    h(t) = t^{3/4} - 3t^{1/4}

    h'(t) = \dfrac{3}{4} u^{-1/4} - (\dfrac{1}{4}) 3 v^{-3/4}

    h'(t) = \dfrac{3}{4} u^{-1/4} - \dfrac{3}{4} v^{-3/4}

    h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}

    h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4} = 0

    h'(t) = -\dfrac{3}{4} t^{-3/4} = -\dfrac{3}{4} t^{-1/4}

    h'(t) = t^{-3/4} = t^{-1/4} = t^{-4/4} = t^{-1}

    You are giving multiple = sign in the same line without showing how you got the right side from the left side. Help us understand what you are doing by writing it. We cannot decipher what you have in mind what you write your expression the way you do. Perhaps you need to write one = sing in a line, and write a short comment next to the line telling what did you do to get from one line to the next.
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