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Math Help - critical points

  1. #1
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    critical points

    The following function on the domain t>0 has a local maximum at t=1/2. True of False?
    <br />
 f(t) = \sqrt t .e^{ - t}   <br />

    When I differentiate I get the following, which I think is correct.
    <br />
 f'(t) = e^{ - t} (\frac{1}{2}t^{\frac{{ - 1}}{2}}  - t^{\frac{-1}{2}} ) <br />

    However I cannot see how t=1/2 is a critical point. I know critical point occur when 1st derivative is equal to zero however, I doesn't appear to me that t=1/2 makes f'(t) zero
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  2. #2
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    Hello,
    Quote Originally Posted by Craka View Post
    The following function on the domain t>0 has a local maximum at t=1/2. True of False?
    <br />
 f(t) = \sqrt t .e^{ - t}   <br />

    When I differentiate I get the following, which I think is correct.
    <br />
 f'(t) = e^{ - t} (\frac{1}{2}t^{\frac{{ - 1}}{2}}  - t^{\frac{-1}{2}} ) <br />

    However I cannot see how t=1/2 is a critical point. I know critical point occur when 1st derivative is equal to zero however, I doesn't appear to me that t=1/2 makes f'(t) zero
    That's because it's not <br />
 f'(t) = e^{ - t} (\frac{1}{2}t^{\frac{{ - 1}}{2}}  - t^{\frac{-1}{2}} ) <br />

    but rather <br />
 f'(t) = e^{-t} (\tfrac 12 t^{\frac{-1}{2}}-t^{\frac{{\color{red}+}1}{2}}) <br />

    f'(t)=e^{-t} (\frac{1}{2 \sqrt{t}}-\sqrt{t})
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  3. #3
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    Thankyou for the reply, I can see where I made a mistake with differentiating. However I can still not see how there is a critical point at t=1/2. Maybe I missing something really elementary but could you point it out.
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  4. #4
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    Quote Originally Posted by Craka View Post
    Thankyou for the reply, I can see where I made a mistake with differentiating. However I can still not see how there is a critical point at t=1/2. Maybe I missing something really elementary but could you point it out.
    \frac{1}{2 \sqrt{t}}-\sqrt{t} = 0

    Multiply both sides by 2 \sqrt{t}:

    1 - 2t = 0 \, ....
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  5. #5
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    f'(t)=e^{-t} (\frac{1}{2 \sqrt{t}}-\sqrt{t})

    0 = (\frac{1}{2 \sqrt{t}}-\sqrt{t})

    0 = \frac{1 -2t}{2\sqrt{t}}

    0 = 1 -2t

    -1 = -2t

    \frac{1}{2} = t
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  6. #6
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    thankyou. Makes sense, only need to consider what was in brackets ie (1/2t^(-1/2)- t^1/2 ). As that will be the only time the function is able to be one as euler to any power is never zero right?
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