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Math Help - Find the absolute minimum

  1. #1
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    Find the absolute minimum

    I need to find the absolute minimum value of f(x)= 8x+ (1/16x^2) on the interval (0, infinity).

    I took the first derivative, f'(x)= 8-(1/8x^3) and set it equal to zero.

    I found that x= 4, but I don't think that's the absolute minimum. Please help!!!
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  2. #2
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    Quote Originally Posted by rust1477 View Post
    I need to find the absolute minimum value of f(x)= 8x+ (1/16x^2) on the interval (0, infinity).

    I took the first derivative, f'(x)= 8-(1/8x^3) and set it equal to zero.

    I found that x= 4, but I don't think that's the absolute minimum. Please help!!!
    May I ask how you got x=4? Check your work: 8-\frac1{8\cdot4^3}=8-\frac1{512}=\frac{4095}{512}\neq0.

    Try again.
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  3. #3
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    Quote Originally Posted by rust1477 View Post
    I need to find the absolute minimum value of f(x)= 8x+ (1/16x^2) on the interval (0, infinity).

    I took the first derivative, f'(x)= 8-(1/8x^3) and set it equal to zero.

    I found that x= 4, but I don't think that's the absolute minimum. Please help!!!
    Go back and re-work - carefully - your solution to f'(x) = 0.
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  4. #4
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    Reworked f'(x)

    I reworked the problem and got (-1/8x^3)=-8 and x=2 but I still don't think that 2 is the absolute minimum value. How should I check it?
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  5. #5
    Senior Member Pinkk's Avatar
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    You're wrong again...

    8-\frac{1}{8x^{3}}=0

    \frac{1}{8x^{3}}=8

    64x^{3}=1

    x^{3}=\frac{1}{64}

    x=\frac{1}{4}
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  6. #6
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    f'(2) does not equal 0
    check again
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