1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    HallsofIvy

    User Avatar
    Staff Emeritus
    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

    Mentallic

    User Avatar
    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.

    [tex]\frac{x}{2}>1+\frac{4}{x}[/tex]

    [tex]\frac{x^3}{2}>\frac{x^2(x+4)}{x}[/tex]

    [tex]x^3>2x^2+8x[/tex]

    [tex]x^3-2x^2-8x>0[/tex]

    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:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Algebra prob
  1. Trig prob (Replies: 8)

  2. Trig identiy prob (Replies: 3)

  3. Limit prob (Replies: 4)

Loading...