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Math Help - More questions:)

  1. #1
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    More questions:)

    1. How would u go about finding the inverse of the following funcions:
    (i can do the regular ones)

    a. f:R\{1} ~ R, f(x)= x+1/x-1
    b. f:R\{2/3}~R, f(x)= 2x+3/3x-2

    2. Let f:S~R be given by f(x)=x-3/2x-1 where S=R\{1/2}
    a. show that ff is defined
    b. Find ff(x)
    c. Write down the inverse of f

    3. Also how would u go about solving ((x-4)/3)^2= x+2 for x?

    Thanks, (ps ~ = arrow)
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  2. #2
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    Quote Originally Posted by chaneliman View Post
    1. How would u go about finding the inverse of the following funcions:
    (i can do the regular ones)

    a. f:R\{1} ~ R, f(x)= x+1/x-1
    b. f:R\{2/3}~R, f(x)= 2x+3/3x-2

    2. Let f:S~R be given by f(x)=x-3/2x-1 where S=R\{1/2}
    a. show that ff is defined
    b. Find ff(x)
    c. Write down the inverse of f

    3. Also how would u go about solving ((x-4)/3)^2= x+2 for x?

    Thanks, (ps ~ = arrow)
    1. a. Let x = (y + 1)/(y - 1). Solve for y and you have the inverse:

    x(y - 1) = y + 1 => xy - x = y + 1 => xy - y = x + 1 => y(x - 1) = x + 1 => y = (x + 1)/(x - 1).

    In other words, f is its own inverse: fof(x) = x. Such a function is called an involution.

    b. Let x = (2y + 3)/(3y - 2). Solve for y and you have the inverse. Proceed as above.

    --------------------------------------------------------------------------------------------

    2. a. What is the necessary condition for fog to be defined? Test this condition for f.

    b. fof(x) = (f - 3)/(2f - 1) = \frac{\frac{x - 3}{2x - 1} - 3}{2\left[ \frac{x - 3}{2x - 1} \right] - 1} = \frac{(x - 3) - 3(2x - 1)}{2(x - 3) - (2x - 1)} = x.

    The penultimate expression is obtained by multiplying top and bottom by (2x - 1).

    c. Use the above result, ie. fof(x) = x.

    ---------------------------------------------------------------------------------------

    3. ((x-4)/3)^2= x+2 => (x - 4)^2 = 9(x + 2). Expand the left hand side, re-arange and simplify to get:

    quadratic = 0.

    Solve the quadratic.
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