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Math Help - Composite Function

  1. #1
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    Composite Function

    hi!

    Attached is a function qns......now studying for a test and come across problems which i have difficulty completing.... thanks for ur help in advance!
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  2. #2
    Grand Panjandrum
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    A function f is defined by:

    f:x\mapsto ax+b,\ x \in \mathbb{R},

    where a and b are positive constants.

    (a) i. Find f^3(x), (where f^3(x)=fff(x)).

    f^2(x) = f(f(f(x))) = f(f(ax+b))


    =f(a(ax+b)+b)
    =a(a(ax+b)+b)+b
    =a^3x+a^2b+ab+b
    (ii) Given that f^3(x)=64x+21, find the values of a and b.

    From part (i) we know that the coefficient of x is a^3, so we have a^3=64. or a=4.

    We also know that the constant term is a^2b+ab+b=21, which when we substitute in the known value of a gives us b=1.

    RonL

    (other parts to follow in another post if someone else does not provide their solutions first).
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by CaptainBlack View Post
    A function f is defined by:

    f:x\mapsto ax+b,\ x \in \mathbb{R},

    where a and b are positive constants.
    (iii) With the values of a and b found in part (ii) determin f^n(x), leaving your answer in the form p^nx+\frac{p^n-1}{q} where p and q are integers to be found.

    From the way that part (i) went I will assume that:

    <br />
f^n(x)=a^nx + a^{n-1}b + \dots +ab + b<br />

    Putting in a=4 and b=1:

    f^n(x)=4^n x + 4^{n+1}+\cdots + 4 + 1

    Which simplifies to:

    <br />
f^n(x)=4^nx + \frac{4^n-1}{3}


    RonL
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  4. #4
    Grand Panjandrum
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    Another function g is defined by g: x\mapsto e^x,\ x \in \mathbb{R},\ x<0.

    (b) (i) Find in terms of a and b the range of fg.

    With the definition of g the range of g is the open interval (0,1), which f will map to the open interval (b, (b+a))

    (ii) If h(x)=[g(x)]^2, determine, with a reason, whether h^{-1} exists.

    With the give devinition of g(x) we have h(x)=e^{2x}, and the range of h is the open interval (0,1). But \ln is defined everywhere on this interval and single valued, so if:

    h(x)=y,

    then x=\ln(y)/2, and is the only such x with image under h of y


    RonL
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  5. #5
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    thanks for ur help !
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