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Annoying inequalities question

  1. Aug 29, 2014 #1
    1.On the same axis sketch the graphs of y = (x-a)^-1 and y = 4|x-a|
    This part i have completed, the first equation has a horizontal asymptote at x = a, it being rectangular hyperbola. the second equation drawn too.

    2. Solve (x-a)^-1 < 4|x-a|, giving your answers in a.
    Now the second part confuses me, because i am not sure where to start as its the first of its kind I've tackled with an arbitrary constant.
    usually to solve something like this i would multiply both sides by the denominator squared, to ensure that i haven't multiplied by a negative number. I have tried equating two equations to y, one having the positive modulus and another having the negative, but in the end i get a really peculiar expression. (and it doesn't match the one in the answer section )

    please help! i haven't started my degree yet (Mathematics). so if my terminology is off could someone please help me out there too haha.
     
  2. jcsd
  3. Aug 29, 2014 #2

    ehild

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    Show your work, please.

    ehild
     
    Last edited: Aug 29, 2014
  4. Aug 29, 2014 #3

    HallsofIvy

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    You shouldn't have any "formula" to deal with at all! The point of "2" after "1" is clearly to use the graph you drew in "1"! Color or shade the region where the graph of 1/(x- a) is below the graph of 4|x- a|.
     
  5. Aug 29, 2014 #4

    Ray Vickson

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    What you call a "horizontal asymptote" I would call a "vertical asymptote". I am speaking here of the graph of y = 1/(x-a).

    To deal with 1/(x-a) < 4|x-a|, look at the two cases (1) x < a; and (2) x > a. Can you see why these two cases are separated, and why it is important to deal with them?
     
    Last edited: Aug 29, 2014
  6. Aug 29, 2014 #5
    Thanks HallsofIvy for your reply, the question is asking me to sketch the graphs of the said functions, this wouldn't yield an accurate value for where the functions intersect or where one is 'less than/more than' the other. The expression has to be solved algebraically.
    Thats Ray Vickson.
    Yes, sill of me, vertical it is!
    The question does not ask to solve
    1/(x-a) < 4|x-1|

    but asks to solve
    1/(x-a) < 4|x-a|

    If it was requesting the former rather than the latter i would be able to solve that relatively easily.
    The fact that the 'a' is encapsulated within the modulus sign is throwing me off.


    this is what i have so far

    (x-a)^-1 < 4|x-a|

    [4(x-a)]^-1 < |x-a|

    now i split this into

    (1).... [4(x-a)]^-1 < x-a and (2).... [4(x-a)]^-1 < -(x-a)
    dealing with (1)

    1 < 4(x-a)^2
    1 < 4(x^2 - 2ax + a^2)
    4x^2 - 8ax + (a^2 + 1) < 0

    Then using the quadratic formula i get this unusual surd thing going on which does not resemble either answer in the back.

    If this is the correct method then there must be a silly error somewhere.
     
  7. Aug 29, 2014 #6

    Ray Vickson

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    Right: |x-1| was a typo; changed to |x-a| now.
     
  8. Aug 29, 2014 #7

    Ray Vickson

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    If x < a, what is the sign of 1/(x-a)? What is the sign of 4|x-a|? What can you say about the desired inequality?

    If x > a, what does the plot of y = 1/(x-a) look like for x only very slightly greater than a? What does the plot of y = 4|x-a| look like in this case? How do the two compare? If x >> a (that is x is much, much larger than a), what does y = 1/(x-a) look like? What does y = 4|x-a| look like? How do they compare? So, now .... ?
     
  9. Aug 29, 2014 #8

    ehild

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    You can introduce y=x-a and solve the inequality 1/y<4|y| for y. Then replace y=x-a.

    That is utterly wrong. It should be 4x^2 - 8ax + (4a^2 - 1) > 0

    ehild
     
  10. Aug 30, 2014 #9

    Ray Vickson

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    You say " ...this wouldn't yield an accurate value for where the functions intersect or where one is 'less than/more than' the other"---meaning that the graph does not give an exact answer. Of course not, but that is not the point. By looking at the graphs and (roughly) where they cross, you can more easily set up meaningful equations for the "boundary points", and you can the go ahead and solve those to as much accuracy as you want. The graph is a very helpful crutch, not the final answer.
     
  11. Aug 30, 2014 #10
    Got up this morn and solved it! Thanks guys.
    The sketch does actually help a lot when choosing which direction to go in!
     
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