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    (Original post by MrToodles4)
    f(x) = 2+x^1/2

    part 1) find an expression for f^-1(x) - I got this: f^-1(X) = (x-2)^2

    part 2). Sketch the graph in the same set of axes the graph of f(x0 and the graph of f^-1(x), clearly marking the line of reflection between the two graphs. - I also got this.

    Part 3). Show that x=4 is the only solution of the equation f(x) = f^-1(x) - This is the bit I do not understand - normally I just take the original function so int his case 2+x^1/2 and equal this to y=x and solve - but the answer says you cannot use this method - I don't understand why so any help will be greatly appreciated. They took f^-1(x) and equalled this to y=x, Why is this the case?! won't both lines (all 3 points) intersect at y=x???
    What do you mean equate to y=x?

    Solve f^-1 (x) = f (x), upon solving you square both sides which causes extra solutions which you have to check in the original equation to see if they hold true.
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    (Original post by NotNotBatman)
    What do you mean equate to y=x?

    Solve f^-1 (x) = f (x), upon solving you square both sides which causes extra solutions which you have to check in the original equation to see if they hold true.
    I got the answer now thank you. I got X = 4 and X =1. The answer says x=4 since x>2 (I think thats the domain of the new f^-1(x) function)
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    (Original post by MrToodles4)
    I got the answer now thank you. I got X = 4 and X =1. The answer says x=4 since x>2 (I think thats the domain of the new f^-1(x) function)
    It's because 1 doesn't work in the original equation  (1-2)^2 \neq 2+\sqrt{1}
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    (Original post by NotNotBatman)
    It's because 1 doesn't work in the original equation  (1-2)^2 \neq 2+\sqrt{1}
    Oh I see - so stuff like this I need to be weary of. Thanks
 
 
 
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