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anonymous

  • one year ago

Find the absolute minimum and absolute maximum values of f on the given interval. f(x) = x − ln 3x, [1/2,2] I got -0.0986.. for the min but i got it wrong, help!

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  1. mathstudent55
    • one year ago
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    f(x) = x - ln 3x f'(x) = 1 - 1/x 1 - 1/x = 0 x - 1 = 0 x = 1 Minimum is at x = 1

  2. anonymous
    • one year ago
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    so my max is wrong too, this is how i was solving it 0.5-ln(1.5) and so on

  3. mathstudent55
    • one year ago
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    Look at the graph of the function. The maximum may be at 0.5 or at 2. Evaluate the function at those two points.

  4. mathstudent55
    • one year ago
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    You evaluated the function at 0.5, but the maximum may be at another point.

  5. anonymous
    • one year ago
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    right, and then i did the same for the min and I got-0.098 but when i typed my answer, it said it was wrong 1-ln(3)

  6. mathstudent55
    • one year ago
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    For the minimum, I get it at x = 1.

  7. mathstudent55
    • one year ago
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    When x = 1, f(1) = 1 - ln 3 = -0.0986

  8. mathstudent55
    • one year ago
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    f(1) = 1 - ln 3 is the exact value of the minimum.

  9. anonymous
    • one year ago
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    yeah and i used -0.0986 for the min, well i rounded so i used -0.10 and it was wrong

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