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Math Help - concavity

  1. #1
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    concavity

    Im having trouble with finding the minimum value.

    the function is (x)(e^5/x)

    my derivative is (e^5/x)(-5x+1)

    I set the derivative to 0 and get
    x=-1/-5 (.2)

    Plugging this back into the original equation gets me
    14400979867.5...which is wrong. Any help would be appriciated
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  2. #2
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     f(x) = x * e^\frac{5}{x}

    Differentiating,

     \frac{df}{dx} = \frac{d}{dx} (x * e^\frac{5}{x}

    Applying the product rule,

     \frac{df}{dx} = e^\frac{5}{x} * \frac{d}{dx} (x) + x * \frac{d}{dx} (e^\frac{5}{x}

    By the chain rule,
     = \frac{df}{dx} = e^\frac{5}{x} + x (e^\frac{5}{x} * \frac{d}{dx}(\frac{5}{x})

    Which simplifies into

     f'(x) = e^\frac{5}{x} - \frac{5 * e^\frac{5}{x}}{x}

    Which can also be written as

     = e^\frac{5}{x} ( \frac{x -5}{x})

    So your derivative was wrong in the first place. Might be why your answer was coming up wrong
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  3. #3
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    Quote Originally Posted by Evan.Kimia View Post
    Im having trouble with finding the minimum value.

    the function is (x)(e^5/x)

    my derivative is (e^5/x)(-5x+1)

    I set the derivative to 0 and get
    x=-1/-5 (.2)

    Plugging this back into the original equation gets me
    14400979867.5...which is wrong. Any help would be appriciated
    Shouldn't the derivative be (e^5/x)(-5x^(-1)+1)?
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  4. #4
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    Hmm... so i tried 5 since x=0 does make y=0 but 0 isnt in the domain. I got 13.59~ for my min value but it didnt work
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  5. #5
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     x = 0

    Actually won't make f'(x) = 0.

     f'(x) = e^\frac{x}{5}(\frac{x - 5}{x})

    Plugging in x = 0,

     f'(0) = 1 (\frac{-5}{0})

    Which is clearly undefined. So, as you've said, x = 5 is your only critical point.

    Now, taking the second derivative,

     f''(x) = \frac{25 * e^\frac{5}{x}}{x^3}

    It is readily verified that this function is only concave upwards, so there is only an absolute min.
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  6. #6
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    y = x e^{\frac{5}{x}}

    y' = x e^{\frac{5}{x}} \cdot \left(-\frac{5}{x^2}\right) + e^{\frac{5}{x}}

    y' =  e^{\frac{5}{x}}\left(1 - \frac{5}{x}\right)

    y' is undefined at x = 0

    for x < 0 , y' > 0 ... there are no extrema in this interval.

    y' = 0 at x = 5

    for 0 < x < 5 , y' < 0

    for x > 5 , y' > 0

    a relative minimum exists at x = 5 , the value of which is 5e.

    The function has no absolute minimum. Consider what happens to the function as x \to -\infty.
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  7. #7
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    Thanks, but when i plug x=5 back into my origional equation to get the min value i get 13.5914091423 which isnt correct
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  8. #8
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    Did you copy the problem down correctly?
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  9. #9
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    grrr... it wanted me to round it to 2 decimal places... thank you.
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