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Conditions for a function to be constant

  1. Oct 28, 2012 #1
    1. The problem statement, all variables and given/known data

    If a, b and c are nonzero distinct real numbers and c > 0. The even function given by:

    f(x) = (ax + b)/(x + c)

    is constant for -c < x < c and it's value is

    (i) a + b; (ii) a + c; (iii) c; (iv) b; (v) a

    2. Relevant equations



    3. The attempt at a solution

    Okay, so first I solved for x but got two different answers because the function is even:

    x = (b - f(x)c)/(f(x) - a) and x = (b - f(x)c)/(a - f(x))

    Then I wanted to check the conditions on f(x) for -c < x < c:

    -c < (b - f(x)c)/(f(x) - a)

    But I don't know how to proceed, because I want to divide both sides by (f(x) - a) and I don't know if it's a negative number or not. What should I do here? How does the information given in the problem statement help me at this point? Am I taking a wrong approach at this problem?
     
  2. jcsd
  3. Oct 28, 2012 #2

    mfb

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    If the function is constant, you cannot solve for x - by definition, all x-values give the same f(x).

    With arbitrary a,b,c, the function will not be constant - you can use the information "function is constant" to find a relation between those parameters. If you can use derivatives, they can give you that relation (hint: what is the derivative of a constant function?). Alternatively, consider x=0 and x=c/2 or similar values. f(x) has to be the same for both x-values.

    If you have that relation and use it in your definition of f(x), you'll get the result.
     
  4. Oct 28, 2012 #3

    ehild

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    You get a relation among a,b,c from the condition that the function is even, that is f(x)=f(-x).

    ehild
     
  5. Oct 28, 2012 #4
    Also, does x < y > z imply z < x < y?

    EDIT: I mean if x < y and y is a variable; does that imply that x is less than the smallest value y can take on?

    I have that if x > 4 than 2 < (x^2)/2 - 6 so if y < (x^2)/2 - 6 does that mean y < 2?
     
  6. Oct 28, 2012 #5

    mfb

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    4 < 6 > 5
    But not 5 < 4 < 6.

    That does not make sense.

    No, unless you have "y < (x^2)/2 - 6 for all x>4"
     
  7. Oct 28, 2012 #6

    Mark44

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    You shouldn't write stuff like x < y > z. Do you mean x < y AND y > z?

    The notation a < b < c is shorthand for a < b AND b < c, and means that b lies between a and c.
     
  8. Oct 28, 2012 #7
    Okay. So if I have to prove something like: if x > 4 and y < 2 then x^2 - 2y > 12.

    Can I say that the statement also implies that if x^2 - 2y > 12 then x > 4 and y < 2? Or that if x^2 - 2y > 12 and x > 4 then y < 2? Maybe I can prove all that but is that implied in the original statement? If not, how can I prove or disprove something like that?
     
  9. Oct 28, 2012 #8

    mfb

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    @V0ODO0CH1LD: Is that related to your original question in any way?

    That is true as x^2 - 2y > 16 - 2*2 = 12

    x=0, y=-7 is a counterexample.

    That statement is true as x^2-2y < 16-2y, and 16-2y > 12 implies y < 2. But you cannot conclude it from the original statement alone.


    "if x > 4 and y < 2 then x^2 - 2y > 12"
    directly implies
    "if not x^2 - 2y > 12, then (not x>4) or (not y<2)"
    which can be re-written as
    "if x^2-2y <= 12, then x<=4 or y>=2"
    and nothing else.
     
  10. Oct 28, 2012 #9
    Yeah, I was just using something else to better exemplify what I was having trouble with. But thanks, you really helped out! Just one thing; is that method you used to prove that if x > 4 and y < 2 then x^2 - 2y > 12 always valid?
     
  11. Oct 28, 2012 #10

    mfb

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    Which method? "Using inequalities"? If you do it right, you can always use them.
     
  12. Oct 28, 2012 #11

    Mark44

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    As mfb said, it's just properties of inequalities.

    If x > 4 and y < 2, then x2 > 16 and 2y < 4 ##\Leftrightarrow## 2y + 12 < 16
    Hence x2 > 2y + 12, or equivalently, x2 - 2y > 12.
     
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