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    x^2 - 2x - 3, x is real, x >/= 1

    Write down the domain and range of f^-1

    Can someone explain this to me? I dont get this function stuff at all..
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    (Original post by imasillynarb)
    x^2 - 2x - 3, x is real, x >/= 1

    Write down the domain and range of f^-1

    Can someone explain this to me? I dont get this function stuff at all..
    I think probably the best thing is to draw the graph and work out the lowest/highest point to determine the range. Don't know if that helps - i have a problem with functions as well
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    yea i hate functions: what does y = / 1/x / look like
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    (Original post by TheWolf)
    yea i hate functions: what does y = / 1/x / look like
    If you mean y = | 1/x |, then it is just reflected on the x-axis below x=0, because y must always be positive.

    However, if you have y = 1/|x|, then below x=0 it is reflected on the y-axis, because the graph acts as if x is positive all the time.
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    (Original post by mik1a)
    If you mean y = | 1/x |, then it is just reflected on the x-axis below x=0, because y must always be positive.

    However, if you have y = 1/|x|, then below x=0 it is reflected on the y-axis, because the graph acts as if x is positive all the time.
    thanks!
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    (Original post by imasillynarb)
    x^2 - 2x - 3, x is real, x >/= 1

    Write down the domain and range of f^-1

    Can someone explain this to me? I dont get this function stuff at all..
    f(x) = x^2 - 2x - 3

    y = f(x)

    Change round:

    x = f(y)

    so if:

    y = x^2 - 2x - 3

    We change to:

    x = y^2 - 2y - 3
    x = (y - 1)^2 - 4
    x + 4 = (y - 1)^2
    +/- root(x+4) = y - 1
    y = 1 +/- root(x+4)

    Therefore:

    f^-1(x) = 1 +/- root(x+4)
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    (Original post by imasillynarb)
    x^2 - 2x - 3, x is real, x >/= 1

    Write down the domain and range of f^-1

    Can someone explain this to me? I dont get this function stuff at all..
    Because this is the inverse of the function f, the ranges and domains will swap round.

    This works because the function is one-one mapping, so only one f(x) value can be found from one x value.

    So the range is clearly x >_ 1.
    The domain:

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

    Showing a min. point of -4.
    The domain is therefore:

    f(x) >_ -4
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    (Original post by mik1a)

    f^-1(x) = 1 +/- root(x+4)
    u ignore the -ve root though as functions and their inverses must be 1 to 1 i think. therefore, you can not have two values of x that make one of y nor two ys that give an x.
    so f^-1(x)=1+ l rt(x+4) l

    domain of inverse is the same as range of normal
    so the range of this will be f^-1(x) >/= 1

    and domain is x>=-4
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    (Original post by mik1a)
    f(x) = x^2 - 2x - 3

    y = f(x)

    Change round:

    x = f(y)

    so if:

    y = x^2 - 2x - 3

    We change to:

    x = y^2 - 2y - 3
    x = (y - 1)^2 - 4
    x + 4 = (y - 1)^2
    +/- root(x+4) = y - 1
    y = 1 +/- root(x+4)

    Therefore:

    f^-1(x) = 1 +/- root(x+4)


    mmmm i drew the graph out, and i know that x is bigger or equal to one. when x = 1, y = -4, and it is also the minimum point. so isnt the domain > or equal to -4?
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    (Original post by mik1a)
    Because this is the inverse of the function f, the ranges and domains will swap round.

    This works because the function is one-one mapping, so only one f(x) value can be found from one x value.

    So the range is clearly x >_ 1.
    The domain:

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

    Showing a min. point of -4.
    The domain is therefore:

    f(x) >_ -4
    Just to clarify... the domain of f(x) is x >= 1
    and the domain of f^-1 (x) is x >= -4 ? is that the same as the range of f(x)?
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    (Original post by kimoni)
    Just to clarify... and the domain of f^-1 (x) is x >= -4 ? is that the same as the range of f(x)?
    yes..it is supposed to be
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    (Original post by lgs98jonee)
    yes..it is supposed to be
    So basically, the range of f^-1 is the domain of f, and the domain of f^-1 is the range of f ?
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    (Original post by imasillynarb)
    So basically, the range of f^-1 is the domain of f, and the domain of f^-1 is the range of f ?
    Heh, it's funny how late on in the year we learn such handy things...
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    (Original post by imasillynarb)
    So basically, the range of f^-1 is the domain of f, and the domain of f^-1 is the range of f ?
    yes
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    (Original post by TheWolf)
    yes
    aaah! that's useful!
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    So when asked to find a range of a function we find its max/min point and say the range is higher than or lower than a specific point? Also, if you were trying to the domain of a function and there was one point where x was an asymptote, how would you express the domain???

    Thanks
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    (Original post by kimoni)
    Heh, it's funny how late on in the year we learn such handy things...
    Yeh...wish my teacher would of said that
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    (Original post by Silly Sally)
    So when asked to find a range of a function we find its max/min point and say the range is higher than or lower than a specific point? Also, if you were trying to the domain of a function and there was one point where x was an asymptote, how would you express the domain???

    Thanks
    eg.

    f(x) = 1/x

    Domain of f(x):

    f(x) < 0
    f(x) > 0
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    (Original post by Silly Sally)
    So when asked to find a range of a function we find its max/min point and say the range is higher than or lower than a specific point? Also, if you were trying to the domain of a function and there was one point where x was an asymptote, how would you express the domain???

    Thanks
    to find range of function, because range is to do with the y values, just draw the graph, and find the x value of minimum or maximum point, and put this x value into the curve's equation to find y value of the minimum/maximum point
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    (Original post by mik1a)
    f(x) = x^2 - 2x - 3

    y = f(x)

    Change round:

    x = f(y)

    so if:

    y = x^2 - 2x - 3

    We change to:

    x = y^2 - 2y - 3
    x = (y - 1)^2 - 4
    x + 4 = (y - 1)^2
    +/- root(x+4) = y - 1
    y = 1 +/- root(x+4)

    Therefore:

    f^-1(x) = 1 +/- root(x+4)
    yes so that means the domain of f^-1(x) > -4 ?



    you could also do it a different way: just complete the square to get
    0=(x-1)^2 -4


    so minima of f(x) is (1,-4) so the range of f(x)>-4,
    therefore domain of f^-1(x)>-4
 
 
 
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