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Math Help - Querry: Relations and functions

  1. #1
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    Post Querry: Relations and functions

    Hello people , I was recently studying relations and functions and encountered some questions that i wasn't able to solve. Please help me solve the following questions.

    1) Find the range of the functions:

    (a) f(x) = 4(x^2) - 4x + 3, x E R
    (b) f(x) + 4(x^2) - 4x + 8, x E R

    2) If f(x) = (x + 3 / 4x - 5) and y= 3 + 5x / 4x -1 , show that f(y) = x

    3) If f(x) = (x - 1 / x + 1); x is not equal to -1, show that f( x -1 / x + 1) = -1 / x

    4) If f(x) = x^2 + 3x -2, -1 ≤ x ≤ 1, find the value of x for which f(x) = f(2x + 1)

    5) If f(x) = x / 2 and g(x) = 2x + 1, show that 2g [f(x)] = 2f[g(x)] + 1
    Last edited by saberteeth; August 25th 2009 at 05:07 AM.
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  2. #2
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    Quote Originally Posted by saberteeth View Post
    Hello people , I was recently studying relations and functions and encountered some questions that i wasn't able to solve. Please help me solve the following questions.

    1) Find the range of the functions:

    (a) f(x) = 4(x^2) - 4x + 3, x E R
    (b) f(x) + 4(x^2) - 4x + 8, x E R

    2) If f(x) = (x + 3 / x - 5) and y= 3 + 5x / 4x -1 , show that f(y) = x

    3) If f(x) = (x - 1 / x + 1); x is not equal to -1, show that f( x -1 / x + 1) = -1 / x

    4) If f(x) = x^2 + 3x -2, -1 ≤ x ≤ 1, find the value of x for which f(x) = f(2x + 1)

    5) If f(x) = x / 2 and g(x) = 2x + 1, show that 2g [f(x)] = 2f[g(x)] + 1
    1) a) f(x) = 4x^2 - 4x + 3

     = 4\left(x^2 - x + \frac{3}{4}\right)

     = 4\left[x^2 - x + \left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right)^2 + \frac{3}{4}\right]

     = 4\left[\left(x - \frac{1}{2}\right)^2 + \frac{1}{2}\right]

     = 4\left(x - \frac{1}{2}\right)^2 + 2.


    Notice that a = 4. Since this is positive, the turning point is a minimum.

    The turning point is (x, y) = \left(\frac{1}{2}, 2\right).

    Therefore the minimum value of y is 2.

    Thus the range is y \geq 2 or y \in [2, \infty).

    Do the same process for 1 b).
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  3. #3
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    3) Replace x with \frac{x-1}{x+1}:

    f\left(\frac{x-1}{x+1}\right)=\frac{\displaystyle\frac{x-1}{x+1}-1}{\displaystyle\frac{x-1}{x+1}+1}=-\frac{1}{x}

    4) x^2+3x-2=(2x+1)^2+3(2x+1)-2

    Expand and solve the quadratic.

    5) g(f(x))=g\left(\frac{x}{2}\right)=2\cdot\frac{x}{2  }+1=x+1

    f(g(x))=f(2x+1)=\frac{2x+1}{2}

    Now prove the equality.
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  4. #4
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    Cool!
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  5. #5
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    I made an error while typing the 2nd problem in the above list which is now corrected. can you answer that too?
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