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C4 solving composite equation help (why reject the positive x value?) Watch

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    I'm stuck on the last question c(ii)

    I got

    Then when I made it equal to 1 I ended up with
    x^2 = 16
    x = 4, x = -4

    Then the mark scheme has rejected x = 4, and made the final answer x = -4.
    I can't grasp why it's done this, the domain for f(x) is 1=<x<=16, and g(x) is -4=<x<=-1.
    So 4 and -4 are included in the domains together, but for some reason the mark scheme only cares about the g(x) domain, I don't know why it has disregarded the f(x) domain, can anyone help?
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    (Original post by Sayless)

    Then the mark scheme has rejected x = 4, and made the final answer x = -4.
    I can't grasp why it's done this, the domain for f(x) is 1=<x<=16, and g(x) is -4=<x<=-1.
    So 4 and -4 are included in the domains together, but for some reason the mark scheme only cares about the g(x) domain, I don't know why it has disregarded the f(x) domain, can anyone help?
    Well, yeah, it's g's domain that matters here.

    Because when you say x=4 and x=-4 is the solution to fg(x) = 1, then you're saying that you can put x=-4 into g(x), this gives you g(-4) and you put that into f(x), to give you f(g(-4)) and that's meant to be 1.

    So you're saying that you take x=4 and you put it into g(x), to give you g(4) and then do f(g(4)) = 1. Can you see what's wrong here? You're putting x=4 into g(x), but that's not allowed.
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    (Original post by Zacken)
    Well, yeah, it's g's domain that matters here.

    Because when you say x=4 and x=-4 is the solution to fg(x) = 1, then you're saying that you can put x=4 into g(x), this gives you g(4) and you put that into f(x), to give you f(g(4)) and that's meant to be 1.

    So you're saying that you take x=-4 and you put it into g(x), to give you g(-4) and then do f(g(-4)) = 1. Can you see what's wrong here? You're putting x=-4 into g(x), but that's not allowed.
    Ah ok so you put the g(x) into the f(x), meaning you only use the domain for g(x), since x = -4 is the only eligible x value to put into g(x). So if it was reveresed and gf(x) you would look at f(x) domain and then put that x value into the gf(x) to solve it, right?
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    (Original post by Sayless)
    Ah ok so you put the g(x) into the f(x), meaning you only use the domain for g(x), since x = -4 is the only eligible x value to put into g(x). So if it was reveresed and gf(x) you would look at f(x) domain and then put that x value into the gf(x) to solve it, right?
    Correct.
 
 
 
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