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Math Help - Finding absolute extrema of function on closed interval

  1. #1
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    Finding absolute extrema of function on closed interval

    Given:

    g(x) = 4(1+\frac{1}{x}+\frac{1}{x^2}), [-4,5]

    [-4,5] being the interval how can I find the absolute extrema of the function on this interval? Step by step is always easiet to follow for me.

    thanks for any help!
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  2. #2
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    Hello, Jimmy!

    This is a very sneaky problem!


    Given: . g(x) \:=\: 4\left(1+\frac{1}{x}+\frac{1}{x^2}\right),\;\;[-4,5]

    Find the absolute extrema of the function on this interval.
    Set the derivative equal to zero . . .

    We have: . g(x) \;=\;4\left(1 + x^{-1} + x^{-2}\right)

    Then: . g'(x) \;=\;4(-x^{-2} - 2x^{-3}) \;=\;0\quad\Rightarrow\quad -4\left(\frac{1}{x^2} + \frac{2}{x^3}\right) \:=\:0

    Multiply by \text{-}\:\!\frac{x^3}{4}\!:\quad x + 2 \:=\:0 \quad\Rightarrow\quad x \:=\:-2
    . . Then: . g(\text{-}2) \:=\:4\left(1 + \frac{1}{\text{-}2} + \frac{1}{(\text{-}2)^2}\right) \:=\:3

    Critical point: . (\text{-}2, 3)


    Test the endpoints of the interval.

    . . g(\text{-}4) \:=\:4\left(1 + \frac{1}{\text{-}4} + \frac{1}{(\text{-}4)^2}\right) \:=\:\frac{13}{4} \:=\:3\tfrac{1}{4}

    . . g(5) \;=\;4\left(1 + \frac{1}{5} + \frac{1}{5^2}\right) \;=\;\frac{124}{25} \;=\;4\tfrac{24}{25}


    I would have written: . \begin{array}{cc}\text{abs. max.} & \left(5,4\frac{24}{25}\right) \\ \\[-4mm] \text{abs. min.} & (-2,3) \end{array}


    Then I saw that x = 0 is a vertical asymptote!

    I found that: . \lim_{x\to0}g(x) \:=\:+\infty\:\text{ from both sides.}
    . . So there is no absolute maximum.

    But we do have an absolute minimum: . (-2,3)

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