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Question Number 151080 by malwan last updated on 18/Aug/21
if (f○f○f○f)(x)=16x+15 find f(x)
$${if}\:\left({f}\circ{f}\circ{f}\circ{f}\right)\left({x}\right)=\mathrm{16}{x}+\mathrm{15} \\ $$$${find}\:{f}\left({x}\right) \\ $$
Answered by Mokmokhi last updated on 18/Aug/21
It is reasonable to say f(x) is linear by rejecting other possibilities. Then f(x)=ax+b for some unknowns a and b. The value of a should be 2 or −2. If a=2. b=1. If a=−2. b=20.
$$\mathrm{It}\:\mathrm{is}\:\mathrm{reasonable}\:\mathrm{to}\:\mathrm{say}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{linear}\:\mathrm{by}\:\mathrm{rejecting}\:\mathrm{other}\:\mathrm{possibilities}. \\ $$$$\mathrm{Then}\:{f}\left({x}\right)={ax}+{b}\:\mathrm{for}\:\mathrm{some}\:\mathrm{unknowns}\:{a}\:\mathrm{and}\:{b}. \\ $$$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:{a}\:\mathrm{should}\:\mathrm{be}\:\mathrm{2}\:\mathrm{or}\:−\mathrm{2}. \\ $$$$\mathrm{If}\:{a}=\mathrm{2}.\:{b}=\mathrm{1}. \\ $$$$\mathrm{If}\:{a}=−\mathrm{2}.\:{b}=\mathrm{20}. \\ $$
Commented by malwan last updated on 18/Aug/21
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Answered by qaz last updated on 18/Aug/21
let f(x)=x ,we get fixed point x=−1 so f(x)=16(x+1)−1 ⇒(f○f)(x)=16^2 (x+1)−1 (f○f○f)(x)=16^3 (x+1)−1 (f○f○f○f)(x)=16^4 (x+1)−1 (f○f○...○f_(n) )(x)=16^n (x+1)−1
$$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}\:,\mathrm{we}\:\mathrm{get}\:\mathrm{fixed}\:\mathrm{point}\:\mathrm{x}=−\mathrm{1} \\ $$$$\mathrm{so}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{16}\left(\mathrm{x}+\mathrm{1}\right)−\mathrm{1} \\ $$$$\Rightarrow\left(\mathrm{f}\circ\mathrm{f}\right)\left(\mathrm{x}\right)=\mathrm{16}^{\mathrm{2}} \left(\mathrm{x}+\mathrm{1}\right)−\mathrm{1} \\ $$$$\:\:\:\:\:\left(\mathrm{f}\circ\mathrm{f}\circ\mathrm{f}\right)\left(\mathrm{x}\right)=\mathrm{16}^{\mathrm{3}} \left(\mathrm{x}+\mathrm{1}\right)−\mathrm{1} \\ $$$$\:\:\:\:\:\:\left(\mathrm{f}\circ\mathrm{f}\circ\mathrm{f}\circ\mathrm{f}\right)\left(\mathrm{x}\right)=\mathrm{16}^{\mathrm{4}} \left(\mathrm{x}+\mathrm{1}\right)−\mathrm{1} \\ $$$$\:\:\:\:\:\:\left(\underset{\mathrm{n}} {\mathrm{f}\circ\mathrm{f}\circ…\circ\mathrm{f}}\right)\left(\mathrm{x}\right)=\mathrm{16}^{\mathrm{n}} \left(\mathrm{x}+\mathrm{1}\right)−\mathrm{1} \\ $$
Commented by malwan last updated on 18/Aug/21
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Commented by qaz last updated on 19/Aug/21
sorry.there is a error. we need make a reverse iteration. (f○f○f○f)(x)=2^4 (x+1)−1 .... f(x)=2(x+1)−1=2x+1
$$\mathrm{sorry}.\mathrm{there}\:\mathrm{is}\:\mathrm{a}\:\mathrm{error}. \\ $$$$\mathrm{we}\:\mathrm{need}\:\mathrm{make}\:\mathrm{a}\:\mathrm{reverse}\:\mathrm{iteration}. \\ $$$$\left(\mathrm{f}\circ\mathrm{f}\circ\mathrm{f}\circ\mathrm{f}\right)\left(\mathrm{x}\right)=\mathrm{2}^{\mathrm{4}} \left(\mathrm{x}+\mathrm{1}\right)−\mathrm{1} \\ $$$$…. \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{2}\left(\mathrm{x}+\mathrm{1}\right)−\mathrm{1}=\mathrm{2x}+\mathrm{1} \\ $$

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