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Math Help - Evaluating a function

  1. #1
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    Red face Evaluating a function

    I am evaluating the function f(x)=x^2-x-lnx

    The derivative should be f'(x)=2x-1-1/x

    I am trying to find the critical numbers:

    0=2x-1-1/x

    Here comes my problem... I used two different methods but my results are different

    Method one:
    0=2x-1-1/x

    0=(2x^2/x)-(x/x)-(1/x) (common denominator)

    0=(2x^2-x-1)/x

    And so on...


    Method two:

    0=2x-1-1/x

    1=2x-1/x

    1=(2x^2)/x-(1/x)

    1=(2x^2-1)/x

    1*x=(2x^2-1)

    x=(2x^2-1)

    0=(2x^2-x-1)

    And so on...


    When I now compare method one with method two I see that 0=(2x^2-x-1)/x is different from 0=(2x^2-x-1). (Why is my result in method two not devided by x?)
    My question is now, why? And what am I doing wrong? :/

    Please help
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  2. #2
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    Re: Evaluating a function

    There is nothing wrong. Both expressions give identical solutions for x.
    Thanks from Duffman and HallsofIvy
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  3. #3
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    Re: Evaluating a function

    Thank you Gusbob I see that you are correct
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  4. #4
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    Re: Evaluating a function

    If I now continue to find the critical numbers (using both method one and two):

    Method one:

    0=(2x^2-x-1)/x

    0=((2x+1)(x-1))/x

    The critical numbers should be:
    x=-1/2
    x=1
    and x=0 (from the denominator)
    ?


    Method two:

    0=(2x^2-x-1)

    0=((2x+1)(x-1))

    The critical numbers should be:
    x=-1/2
    x=1
    ?


    My final question is.. why is there 3 critical numbers using method one but only 2 critical numbers using method two? Am I thinking worng?
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  5. #5
    Junior Member astartleddeer's Avatar
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    Re: Evaluating a function

    Only the numerator will make the quotient equate to zero. Division by zero is undefined so you can neglect the denominator.

    So  \frac{2x^2 - x -1}{x} = 0 reduces to  (2x + 1)(x - 1) = 0
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  6. #6
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    Re: Evaluating a function

    No, you cannot "neglect the denominator" for this problem. astarledeer is correct that a fraction, and so the dervative, is 0 only when the numerator is 0, independent of the denominator. But this problem asked for "critical points" which are points at which the derivative is 0 or where the derivative does not exist. Going back to the original equation,
    f'= 2x- 1- 1/x, f' does not exist at x= 0. However, now we note that the original function, f(x)= x^2- x- ln(x) is also not defined at x= 0. Therefore, the only critical points in the domain of the original function are x= 1/2 and x= 1. But there are examples where the function exists and some point but the derivative does not: y= \sqrt{x} is one.
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  7. #7
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    Re: Evaluating a function

    Duffman
    Mathematics is not just doing additions or subtractions and ..other operations but rather thinking and reasoning.
    first of all do you know what a critical point is?
    if not go here to revise.... Critical point (mathematics) - Wikipedia, the free encyclopedia

    in your case the critical points are 3 as you said oint zero where the derivative does not exist.....this is obvious from the domain of your function...and points 1 and (-1/2)??? but you miss the point again..... the value x= -1/2 is out of the domain of your function...in other words your function is defined in the domain (0,+infinity)..
    and x = -1/2 is not included....therefore the only available critical point to consider is the point x=1.

    MINOAS
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  8. #8
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    Re: Evaluating a function

    Sorry Hall
    your post was not there when I started mine.
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