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Math Help - Critical values and max/mins

  1. #1
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    Critical values and max/mins

    This isn't quite homework, but still stuff I don't understand.

    1- f(x) = 4x^2
    0 < x ≤ 8

    I understand that there's no local max, and that the absolute max is 256. But for local and absolute minimums I don't quite understand. How to find the min values. If I find the critical value and don't include 0, then there's no critical value, which means I don't understand how to find the mins.

    Same thing with this one:

    g(t) = | 4t - 9 |

    g'(t) is 4. Which, as far as I know, doesn't mean there's a critical point.

    One more question:

    x^(-4)*ln(x)

    Product rule. Right. But isn't the derivative of the left hand function x^(-4)*ln(-4), which doesn't exist because you can't take the log of a negative number?
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  2. #2
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    Quote Originally Posted by CFem View Post
    This isn't quite homework, but still stuff I don't understand.

    1- f(x) = 4x^2
    0 < x ≤ 8

    I understand that there's no local max, and that the absolute max is 256. But for local and absolute minimums I don't quite understand. How to find the min values. If I find the critical value and don't include 0, then there's no critical value, which means I don't understand how to find the mins.

    Same thing with this one:

    g(t) = | 4t - 9 |

    g'(t) is 4. Which, as far as I know, doesn't mean there's a critical point.

    One more question:

    x^(-4)*ln(x)

    Product rule. Right. But isn't the derivative of the left hand function x^(-4)*ln(-4), which doesn't exist because you can't take the log of a negative number?
    For the first, there's no absolute min.
    Second, critical points are where g' = 0 or DNE. Here, it's at t = \frac{9}{4}. (BTW -  g' \ne 4)

    Third, not sure how you got the critical point of x = -4. The derivative is

     <br />
\frac{1 - 4 \ln x}{x^5}<br />
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  3. #3
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    Quote Originally Posted by Danny View Post
    For the first, there's no absolute min.
    Second, critical points are where g' = 0 or DNE. Here, it's at t = \frac{9}{4}. (BTW -  g' \ne 4)

    Third, not sure how you got the critical point of x = -4. The derivative is

     <br />
\frac{1 - 4 \ln x}{x^5}<br />
    For the first, does no absolute min imply that there's no local min?

    What am I missing on the second derivative? The derivative of 4t is 4, the derivative of 9 is 0, no?

    For the third one, could you show me your steps? I did this: (I was backwards in my first post)

    F(x) = x^(-4)*ln(x)
    f(x) = x^(-4)
    f'(x) = -4x^(-5)
    g(x) = ln(x)
    g'(x) = 1/x

    F'(x) = f'(x)g(x) + f(x)g'(x)
    F'(x) = -4x^(-5) * ln(x) + \frac{\frac{1}{x^4}}{x}
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  4. #4
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    Quote Originally Posted by CFem View Post
    For the first, does no absolute min imply that there's no local min?

    What am I missing on the second derivative? The derivative of 4t is 4, the derivative of 9 is 0, no?

    For the third one, could you show me your steps? I did this: (I was backwards in my first post)

    F(x) = x^(-4)*ln(x)
    f(x) = x^(-4)
    f'(x) = -4x^(-5)
    g(x) = ln(x)
    g'(x) = 1/x

    F'(x) = f'(x)g(x) + f(x)g'(x)
    F'(x) = -4x^(-5) * ln(x) + \frac{\frac{1}{x^4}}{x}
    1) In general, abs. min doesn't imply there's no rel. min.
    2) You have an absolute function

    g(t) = |4t-9| = \left\{\begin{array}{cl} 4t-9, & t \ge 9/4\\ -(4t-9),& t < 9/4\end{array}\right.

    3) Your answer simplifies to mine.
    Last edited by Jester; October 11th 2009 at 02:35 PM. Reason: fixed typo
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  5. #5
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    Woops, algebra comes back to bite me again.

    Last question I have is for number 1; IS there a local minimum? I would have assumed that since there wasn't an absolute minimum there wouldn't be a local minimum.
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  6. #6
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    Quote Originally Posted by CFem View Post
    Woops, algebra comes back to bite me again.

    Last question I have is for number 1; IS there a local minimum? I would have assumed that since there wasn't an absolute minimum there wouldn't be a local minimum.
    No local min.
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