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Questions related to Relations and Functions

  1. Mar 25, 2015 #1
    1. The problem statement, all variables and given/known data

    1. Range of the function ## \sqrt {x^2+x+1} ## is equal to?

    2.ƒ:R---->R is defined as ƒ(x) = x2 -3x +4, then f -1 (2) is equal to?


    2. Relevant equations
    NA

    3. The attempt at a solution
    For the first one tried squaring on both the sides but that does not give linear x in terms of y for finding the range.

    For second one, I can directly substitute f(x) as 2 for getting the answer.
    But I have a confusion that a function must be injective and surjective for the inverse otherwise it inverse must not exist.
     
  2. jcsd
  3. Mar 25, 2015 #2
    1. you must have ##x^2 + x + 1 \geq 0## so for what x is that true?

    2. I think you could start with ##2 = x^2 - 3x + 4## & go from there
     
  4. Mar 25, 2015 #3
    1. I'm not asking for domain but range.
    I know that the domain would be set of all real numbers.

    2. That I know but I think f inverse should not exist as the function is not one one and onto?
     
  5. Mar 25, 2015 #4

    LCKurtz

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    You have a square root to worry about, so not every ##x## works in the domain. And what ##x##'s work determine the range.

    Sometimes when ##f^{-1}## does not exist, the notation such as ##f^{-1}(2)## means the set of all ##x## such that ##f(x)=2##.
     
  6. Mar 25, 2015 #5
    But I see here that if we choose any x here in this case , the value of
    x2 + x + 1 is always greater than zero,
    so not to worry for square root in this case.
     
  7. Mar 25, 2015 #6

    LCKurtz

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    That's right, but that does not mean the range is ##(0,\infty)##. For example, what ##x## gives ##f(x) = 1/2##?
     
  8. Mar 25, 2015 #7
    No x gives that value
     
  9. Mar 25, 2015 #8

    LCKurtz

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    Right. So you still have to figure out the range.
     
  10. Mar 25, 2015 #9
    Ya, the range then must be greater then 1/2 also.
    What is the method to find the particular value?
     
  11. Mar 25, 2015 #10

    LCKurtz

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    You have to figure out the least that ##x^2+x+1##, and hence its square root, can be.
     
  12. Mar 25, 2015 #11
    I differentiated it,
    Got 2x + 1 = 0
    Hence x = -1/2, a minima.
    Substituting in function we get square root of 3/4 which is √3/2
    Hence range is [√3/2,∞)
    Thanks.
    The word least provoked the differentiation.
     
  13. Mar 25, 2015 #12

    LCKurtz

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    Good. Note that you could have also found the min value without calculus by completing the square.
     
  14. Mar 25, 2015 #13
    Hmm that's also fine.
    As the thread is related to relations and functions.
    I wanted to ask only a last question.

    If ## f(x) = sin^2x + sin^2(x+ π/3) + cosxcos(x+ π/3)## and ##g(5/4)=1##, then ##(gof)(x)## is equal to?
    Options are
    0
    1
    sinx
    None of these

    I know gof(x) is g(f(x)) but here g(x) is not given.
     
  15. Mar 25, 2015 #14
    Anybody there or should I start a new thread?
     
  16. Mar 25, 2015 #15

    SammyS

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    I say yes - new thread. It's quite a different question.

    Composition in LaTeX is ##(f\circ g)(x) \ \ \ \leftarrow\ \ \ \text{(f\circ g)(x)}## .
     
  17. Mar 26, 2015 #16

    Mark44

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    Yes, new thread, which you already did.

    See my reply in that thread.
     
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