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Math Help - First derivative test, turning points

  1. #1
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    First derivative test, turning points

    I don't understand how the LHS equals the RHS for this:

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

    Then this statement follows which i also don't understand.

    "Turning points when dy/dx = 0 are when x^2 - 1 = 0, that is when x = -1, 1.

    Note that dy/dx is undefined at x = 0."


    Any help would be much appreciated.
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  2. #2
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    Quote Originally Posted by anon_404 View Post
    I don't understand how the LHS equals the RHS for this:

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

    Then this statement follows which i also don't understand.

    "Turning points when dy/dx = 0 are when x^2 - 1 = 0, that is when x = -1, 1.

    Note that dy/dx is undefined at x = 0."


    Any help would be much appreciated.
    x^{-2} = \frac{1}{x^2}

    To get the LHS to equal the RHS multiply the 1 by x^2 and combine with the same denominator:

    1 \times \frac{x^2}{x^2} - \frac{1}{x^2} = \frac{x^2-1}{x^2}

    To find the turning points set your equation above to 0 and solve. I assume the above equation is the derivative otherwise that statement is false.

    It is undefined at x=0 because you can't divide by 0 and the x^2 on the bottom would be 0
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  3. #3
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    Quote Originally Posted by anon_404 View Post
    I don't understand how the LHS equals the RHS for this:

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

    Then this statement follows which i also don't understand.

    "Turning points when dy/dx = 0 are when x^2 - 1 = 0, that is when x = -1, 1.

    Note that dy/dx is undefined at x = 0."


    Any help would be much appreciated.

    Hey in terms of the first part:

    x^(-2) is the same as 1/x^2. (RULES OF INDICES)

    So on the LHS you have:

    1- (1/x^2).

    So you have:

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


    Sorry this answer is a bit crap. I'm not that good at explaining things.
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  4. #4
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    Quote Originally Posted by e^(i*pi) View Post

    To get the LHS to equal the RHS multiply the 1 by x^2 and combine with the same denominator:

    1 \times \frac{x^2}{x^2} - \frac{1}{x^2} = \frac{x^2-1}{x^2}

    Would you be able to explain the fraction part more. Sorry i have little algebra knowledge
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  5. #5
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    Quote Originally Posted by anon_404 View Post
    I don't understand how the LHS equals the RHS for this:

    1 - x^(-2) = (x^2 - 1)/x^2
    The RHS is \frac{x^2- 1}{x^2}= \frac{x^2}{x^2}- \frac{1}{x^2}
    = 1- \frac{1}{x^2}= 1- x^{-2}

    Then this statement follows which i also don't understand.

    "Turning points when dy/dx = 0 are when x^2 - 1 = 0, that is when x = -1, 1.

    Note that dy/dx is undefined at x = 0."


    Any help would be much appreciated.
    "Turning points" of a graph must occur where the derivative is 0. A fraction is 0 only where the numerator of is 0. That is, f'(x)= \frac{x^2- 1}{x^2}= 0 when x^2- 1= 0 or x^2= 1 so x= 1 or x= -1.

    Of course, division by 0 is undefined so f'(x)= \frac{x^2-1}{x^2} is undefined when the denominator, x^2, is 0: x^2= 0 so x= 0.
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