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    If f(x)=ax+b and f^(3)(x)=64x+21 find the values of a and b. Suggest a rule for f^(n)(x)

    Would f^3(x) be a^3x+3b and f^n(x)= 4^n(x)+7n?


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    (Original post by Mr Pussyfoot)
    If f(x)=ax+b and f^(3)(x)=64x+21 find the values of a and b. Suggest a rule for f^(n)(x)

    Would f^3(x) be a^3x+3b and f^n(x)= 4^n(x)+7n?


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    It certainly fits so i believe it's correct
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    (Original post by Mr Pussyfoot)
    If f(x)=ax+b and f^(3)(x)=64x+21 find the values of a and b. Suggest a rule for f^(n)(x)

    Would f^3(x) be a^3x+3b and f^n(x)= 4^n(x)+7n?


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    No.

    f(x) = ax+b

    So, f^2(x) = f(f(x)) = a(ax+b)+b

    and

    f^3(x) = f(f(f(x))) = a(a(ax+b)+b)+b

    And take it from there.
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    It may be useful to prove your suggested rule by induction, quite easy.
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    (Original post by ghostwalker)
    No.

    f(x) = ax+b

    So, f^2(x) = f(f(x)) = a(ax+b)+b

    and

    f^3(x) = f(f(f(x))) = a(a(ax+b)+b)+b

    And take it from there.
    Would the rule be a^(n)x+b(a^(n-1)+a^(n-2)...a^0)?




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    (Original post by Mr Pussyfoot)
    Would the rule be a^(n)x+b(a^(n-1)+a^(n-2)...a^0)?
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    Looks good. Notice that the constant term is imply "b" times a G.P., so can be simplified.
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    (Original post by ghostwalker)
    Looks good. Notice that the constant term is imply "b" times a G.P., so can be simplified.
    I see, so the fully simplified version would come down to a^nx+b(1-a^n/1-a) I'm guessing


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    And on another note, if a f(x)=1/(1-x), would f^2(x)= 1/(2+x)? Nevermind, this is incorrect lol


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    (Original post by Mr Pussyfoot)
    And on another note, if a f(x)=1/(1-x), would f^2(x)= 1/(2+x)? Nevermind, this is incorrect lol


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    No, it wouldn't be. It would be \displaystyle f(f(x) = \frac{1}{1 - f(x)} = \frac{1}{1 - \frac{1}{1-x}} = \frac{1-x}{1-x - 1} = \cdots
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    Name:  ImageUploadedByStudent Room1469191557.462888.jpg
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    I've gotten this far on question 15 but have no idea what to do next, should I substitute the values of x into f^-1(x)=f^2(x)?



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    (Original post by Zacken)
    No, it wouldn't be. It would be \displaystyle f(f(x) = \frac{1}{1 - f(x)} = \frac{1}{1 - \frac{1}{1-x}} = \frac{1-x}{1-x - 1} = \cdots
    Yeah I realised this after a while


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    (Original post by Mr Pussyfoot)
    I've gotten this far on question 15 but have no idea what to do next, should I substitute the values of x into f^-1(x)=f^2(x)?
    Nope, nope, nope. This is why I always encourage writing f(f(x)) = \frac{a}{f(x)} + b = \frac{a}{\frac{a}{x} + b} +b \neq \frac{a}{\frac{a}{x}} + b + b
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    (Original post by Mr Pussyfoot)
    I see, so the fully simplified version would come down to a^nx+b(1-a^n/1-a) I'm guessing


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    That's correct.

    Though you can of course simplify further by putting in the actual values of a and b.

    Then as a check set n=3, and see if if gives the desired result. You can also check n=1, to see it gives ax+b, for whatever your values of a,b are.
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    (Original post by Zacken)
    Nope, nope, nope. This is why I always encourage writing f(f(x)) = \frac{a}{f(x)} + b = \frac{a}{\frac{a}{x} + b} +b \neq \frac{a}{\frac{a}{x}} + b + b
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    Okay I arrived at this quadratic, should I follow on to solve for x? I notice that the coefficient of x^2 is similar to what the question is asking for, is there a link?


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    (Original post by Mr Pussyfoot)
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    Okay I arrived at this quadratic, should I follow on to solve for x? I notice that the coefficient of x^2 is similar to what the question is asking for, is there a link?


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    You want the quadratic to be satisfied for all values of x, that is, you need all of the coefficients to be equal to 0 so that the whole thing is equal to 0, i.e: solving for x will give two values of x in terms of a and b such that the quadratic holds, you want the quadratic to be identically 0 for all x (think in terms of: you want it to be an identity not an equation), so anyway.

    The quadratic is \displaystyle (a+b^2)x^2 - b(a+b^2)x - a(a + b^2) = 0 \iff (a + b^2)(x^2 - bx - a) = 0

    So you need all of a+b^2 = 0.
 
 
 
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