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Math Help - HELP! Inequality!

  1. #1
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    HELP! Inequality!

    i have no problem solving inequalities but this one has me stumped, one over x is greater than x. For what real number is 1/x > x ?
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  2. #2
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    Do some manipulation:

    1/x > x (multiply both sides by 1/x)

    1 > x^2 (take the square root of both sides)

    +, - 1 = x

    Check intervals:

    - infinity to - 1

    (-2)

    1/-2 > -2 YES


    -1 to 1

    1/2 works, but -1/2 doesn't...hmmm

    1/(1/2) > 1/2

    2 > 1/2 YES

    1/(-1/2) > -1/2

    -2 > -1/2 NO


    1 to infinity

    (2)

    1/2 > 2 NO


    Solution

    X < -1

    0 < x < 1
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  3. #3
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    Here is a second way to get that solution set.
    \frac{1}{x} > x\; \Rightarrow \;\frac{{1 - x^2 }}<br />
{x} > 0

    The soution set clear.
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  4. #4
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    Plato, doesn't that proceed to the same point?

    (1-X^2)/x >0

    1 - x^2 > 0

    1 > x^2

    etc.
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  5. #5
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    Quote Originally Posted by fcabanski View Post
    Plato, doesn't that proceed to the same point?
    Well yes it is. And I clearly said so.
    But it is a far better way to proceed in that there is no squaring it deal with.
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  6. #6
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    I agree.
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  7. #7
    Super Member dhiab's Avatar
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    Quote Originally Posted by fcabanski View Post
    Plato, doesn't that proceed to the same point?

    (1-X^2)/x >0

    1 - x^2 > 0

    1 > x^2

    etc.
    Hello :This solution is Right if x positif
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  8. #8
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    A very general way of solving complicated inequalities: solve the equation first.

    To solve \frac{1}{x}> x, first look at \frac{1}{x}= x. That's easy: multiplying both sides by x, \frac{1}= x^2 which has roots x= \pm 1. The point of that is that functions can change form ">" to "<" only where they are "=" or where they are not continuous. The functions involved here are equal at x= 1 or x= -1 and not continuous at x= 0 where the left side is not defined. Now we need only check on point in each of the intervals (-\inf, -1), (-1, 0), (0, 1), and (1, \infty).

    For example:
    -2< -1 and if x= -2, \frac{1}{x}= -\frac{1}{2}> -2= x.
    -2< -\frac{1}{2}< 0 and if x= -\frac{1}{2}, \frac{1}{x}= -2< -\frac{1}{2}= x.
    0< \frac{1}{2}< 1 and if x= \frac{1}{2}, \frac{1}{x}= 2> \frac{1}{2}= x
    1< 2 and if x= 2, \frac{1}{x}= \frac{1}{2}< 2= x

    That is, \frac{1}{x}> x for x< -1 or 0< x< 1.
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  9. #9
    No one in Particular VonNemo19's Avatar
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    This same question was asked here.

    http://www.mathhelpforum.com/math-he...questions.html
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  10. #10
    Super Member dhiab's Avatar
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    Hello : THIS THE SOLUTION



    remark : all solution set is open in 0 ; -1; 1, -infinity
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  11. #11
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    Quote Originally Posted by allenfarmer View Post
    i have no problem solving inequalities but this one has me stumped, one over x is greater than x. For what real number is 1/x > x ?
    We are told:

    \frac{1}{x} > x.


    First, we have to note that x \neq 0.


    If we are trying to solve this inequality, we need to take into account two possibilities: x > 0 and x < 0.


    Case 1: x > 0.

    \frac{1}{x} > x

    1 > x^2

    x^2 < 1

    |x| < 1

    -1 < x < 1.


    But since x > 0

    0 < x < 1.



    Case 2: x < 0

    \frac{1}{x} >x

    1 < x^2

    x^2 > 1

    |x| > 1

    x < -1 or x > 1.

    But since x < 0 we have x < -1.


    So putting everything together:

    x < -1 or 0 < x < 1.
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