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Homework Help: Algebra prob

  1. Jul 13, 2009 #1
    1. The problem statement, all variables and given/known data

    (A) Graph the functions f(x)=x/2 and g(x)=1+(4/x) together to indentify the values of x for which x/2>1+(4/x).

    (B) Confirm your findings in part (A) using algebra.

    2. Relevant equations

    3. The attempt at a solution

    (A) I graphed the equations and found the answer to be (-2,0)U(4,infinity). The answer is given in the book and confirms this.

    (B) x/2 > 1+(4/x),

    = x > 2+(8/x)

    = (x^2) > 2x+8

    = (x^2)-2x-8 > 0

    = (x-4)(x+2) > 0

    So doesn't this imply x/2 > 1+(4/x) on the interval (-infinity, -2)U(4, infinity)? Thanks for clearing this up!
  2. jcsd
  3. Jul 13, 2009 #2
    You'll have to be very careful here.

    When you have something along the lines of:

    x > y

    What happens if you multiply both sides by a negative number? This is the same problem you have currently, except with your problem, you will have to consider two cases, one case when x is positive, and another when x is negative, (since you're multiplying through by x) and you will have to deal with the inequality for each case.
    Last edited: Jul 13, 2009
  4. Jul 13, 2009 #3
    Ok, first you need to state here x > 2+(8/x) that x[itex]\neq[/itex]0
    Then you need to solve (x-4)(x+2) > 0

    What should be (x-4) and (x+2) so that the whole equation is > 0 i.e positive?
  5. Jul 13, 2009 #4
    Дьявол, it's true he should also state that [tex]x \neq 0[/tex], but like I said above, in order to get the full answer, he will have to consider two separate cases for when both [tex] x < 0 [/tex] and [tex] x > 0 [/tex].
  6. Jul 13, 2009 #5


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    Science Advisor

    The simplest way to solve an inequality like x/2> 1+ 4/x is to look first at the equality x/2= 1+ 4/x. Now, multiplying on both sides by 2x, [itex]x^2= 2x+ 8[/itex] or [itex]x^2- 2x- 8= (x+ 2)(x- 4)= 0[/itex] which has roots -2 and 4.

    The point is that the only way "f(x)> g(x)" can change to "f(x)< g(x)" is where "f(x)= g(x)" or at points where one or the other of the functions is not continuous. Here the functions are equal at -2 and 4 and 1+ 4/x is not continuous at x= 0.

    If x= -4< -2, x/2= -2 and 1+ 4/x= 1- 1= 0. Yes, -2< 0 so the inequality is satisfied for all x< -2. If x= -1, which is between -2 and 0, x/2= -1/2 and 1+ 4/x= -3. -1/2 is NOT less than -3 so the inequality is NOT satisfied for all x between -2 and 0. You check for x between 0 and 4, and for x> 4.
  7. Jul 16, 2009 #6


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    Homework Helper

    Yes, or...

    if multiplying by x is a problem because if x is negative, it would switch the inequality sign around. Well, how about multiplying by x2? That value is positive no matter if x is positive or negative so there is no need to switch the sign.





    This is a cubic but can be reduced into a quadratic if x is factored out. All you need to do is sketch the cubic and find the intervals that are above the x-axis (greater than 0). You should get the same answer :smile:
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