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Math Help - Critical numbers

  1. #1
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    Critical numbers

    Need a bit of help

    Determine

    a.) All the critical numbers

    b.) all relative extrema

    y(x)=(x-1)e^{2x}

    I'm thinking my derivative is ...

    2*(x-1)*(1)*e^{2x}

    Multiple choice:

    a.){ \frac{1}{2},max}

    b.){0,min}

    c.) { \frac{1}{2}, min}

    d.) none of these

    I'm thinking it is \frac{1}{2} min
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by XIII13Thirteen View Post
    Need a bit of help

    Determine

    a.) All the critical numbers

    b.) all relative extrema

    y(x)=(x-1)e^{2x}

    I'm thinking my derivative is ...

    2*(x-1)*(1)*e^{2x}

    Multiple choice:

    a.){ \frac{1}{2},max}

    b.){0,min}

    c.) { \frac{1}{2}, min}

    d.) none of these

    I'm thinking it is \frac{1}{2} min
    you derivative is wrong. you need the product rule here.
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  3. #3
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    (x-1)2e^{2x}+(1)e^{2x}?
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  4. #4
    Moo
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    Hello,

    Yes, it's it

    (x-1)2e^{2x}+(1)e^{2x}=e^{2x}(2x-2+1)=e^{2x}(2x-1)
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by XIII13Thirteen View Post
    (x-1)2e^{2x}+(1)e^{2x}?
    very good. and Moo was kind enough to factorize it for us. now set Moo's factorized form to zero and solve for your critical points.
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  6. #6
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    Quote Originally Posted by Jhevon View Post
    very good. and Moo was kind enough to factorize it for us. now set Moo's factorized form to zero and solve for your critical points.
    Yes, x=1/2 will zero out the equation. I do believe X>0 is a minimum? If so than c. Thanks guys
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