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    I'm stuck on this domain and range question. I dont know where to start

    The functions f and g are defined by
    f:x→1−ax, x∈ ,
    g:x→x2 +2ax+2, x∈ ,
    where a is a constant.

    (a) Find the range of g in terms of a.
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    Which paper and question?
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    (Original post by Lilly1234567890)
    I'm stuck on this domain and range question. I dont know where to start

    The functions f and g are defined by
    f:x→1−ax, x∈ ,
    g:x→x2 +2ax+2, x∈ ,
    where a is a constant.

    (a) Find the range of g in terms of a.
    Perhaps a sketch of the quadratic may shed some light - you should be able to identify the minimum/maximum points from that and hence determine the range of values g spans.

    NB: It may help you to sketch it if you notice that g(x) = (x+ a)^2 + 2 - a^2
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    [QUOTE=jessedoell;60454181]Which paper and question?[/QUOTE

    (Original post by Zacken)
    Perhaps a sketch of the quadratic may shed some light - you should be able to identify the minimum/maximum points from that and hence determine the range of values g spans.

    NB: It may help you to sketch it if you notice that g(x) = (x+ a)^2 + 2 - a^2
    Thanks for that! I get it now
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    [QUOTE=Lilly1234567890;60455833]
    (Original post by jessedoell)
    Which paper and question?[/QUOTE



    Thanks for that! I get it now
    Great!
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    [QUOTE=Zacken;60455999]
    (Original post by Lilly1234567890)

    Great!
    Hey! Its me again, is it alright if you could help me on this question from the same paper.

    (b) Find an expression for the inverse function f^−1(x) and state its domain.

    F(x)= 3x+2/ x-l

    I was able to find out the inverse which is: f(x)=x+2/x-3

    However, I dont understand how to get the domain. I was thinking of sketching that and finding where it crosses the x-axis but then how on earth do we sketch that graph.
    As you can tell domain and range isn't really my best topic.
    Again, Thanks for the help!
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    [QUOTE=Lilly1234567890;60456171]
    (Original post by Zacken)

    Hey! Its me again, is it alright if you could help me on this question from the same paper.

    (b) Find an expression for the inverse function f^−1(x) and state its domain.

    F(x)= 3x+2/ x-l

    I was able to find out the inverse which is: f(x)=x+2/x-3

    However, I dont understand how to get the domain. I was thinking of sketching that and finding where it crosses the x-axis but then how on earth do we sketch that graph.
    As you can tell domain and range isn't really my best topic.
    Again, Thanks for the help!
    The domain of the inverse function is the range of the function.
    The range of the inverse function is the domain of the function.
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    so the full question is
    f(x)=1+ 4x/ 2x−5 - 15/ 2x^2 −7x+5 , x∈ , x<1.
    (a) show that f(x)= 3x+2/x−1

    (b) Find an expression for the inverse function f −1(x) and state its domain.

    So we need to find the domain of the inverse function which is the same as the range of the function.
    If the domain of the normal function is x<1, to find the range do we sub in X=1 to the function to find y and from that we can deduce the range?
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    (Original post by Lilly1234567890)
    so the full question is
    f(x)=1+ 4x/ 2x−5 - 15/ 2x^2 −7x+5 , x∈ , x<1.
    (a) show that f(x)= 3x+2/x−1

    (b) Find an expression for the inverse function f −1(x) and state its domain.

    So we need to find the domain of the inverse function which is the same as the range of the function.
    If the domain of the normal function is x<1, to find the range do we sub in X=1 to the function to find y and from that we can deduce the range?
    In this case, it's easier to just look at f^{-1}(x) = \dfrac{x+2}{x-3} and see that the denominator can't be 0 so x \in \mathbb{R}, x \neq 3 is the domain.
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    (Original post by Zacken)
    In this case, it's easier to just look at f^{-1}(x) = \dfrac{x+2}{x-3} and see that the denominator can't be 0 so x \in \mathbb{R}, x \neq 3 is the domain.
    Thanks for the help!
    The answer was x<3, XER though.
 
 
 
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