Question-217417

Question Number 217417 by peter frank last updated on 13/Mar/25

Commented by mr W last updated on 13/Mar/25

for f(g(x))=(x^2 +1)^(50) there is no unique solution for f(x) and f(x).

$${for}\:{f}\left({g}\left({x}\right)\right)=\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{50}} \:{there}\:{is}\:{no} \\ $$$${unique}\:{solution}\:{for}\:{f}\left({x}\right)\:{and}\:{f}\left({x}\right). \\ $$

Answered by SdC355 last updated on 13/Mar/25

(f○g)(x)=(x^2 +1)^(50) we can renote (f○g)(x) as f(g(x)) f(g(x))=(x^2 +1)^(50) but we can′t find f(x) and g(x)..... because it dosen′t matter if f(x)=x^(50) and g(x)=x^2 +1 and f(x)=x^(25) and g(x)=(x^2 +1)^2 So we can′t Find a specific function f and g

$$\: \\ $$$$\left({f}\circ\mathrm{g}\right)\left({x}\right)=\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{50}} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{renote}\:\left({f}\circ\mathrm{g}\right)\left({x}\right)\:\mathrm{as}\:{f}\left(\mathrm{g}\left({x}\right)\right) \\ $$$${f}\left(\mathrm{g}\left({x}\right)\right)=\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{50}} \\ $$$$\mathrm{but}\:\mathrm{we}\:\mathrm{can}'\mathrm{t}\:\mathrm{find}\:{f}\left({x}\right)\:\mathrm{and}\:\mathrm{g}\left({x}\right)….. \\ $$$$\mathrm{because}\:\mathrm{it}\:\mathrm{dosen}'\mathrm{t}\:\mathrm{matter}\:\mathrm{if} \\ $$$${f}\left({x}\right)={x}^{\mathrm{50}} \:\mathrm{and}\:\mathrm{g}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{1}\:\mathrm{and} \\ $$$${f}\left({x}\right)={x}^{\mathrm{25}} \:\mathrm{and}\:\mathrm{g}\left({x}\right)=\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{So}\:\mathrm{we}\:\mathrm{can}'\mathrm{t}\:\mathrm{Find}\:\mathrm{a}\:\mathrm{specific}\:\mathrm{function} \\ $$$${f}\:\mathrm{and}\:\mathrm{g} \\ $$

Answered by MathematicalUser2357 last updated on 13/Mar/25

$${f}\left({x}\right)={x}^{\mathrm{50}} \\ $$$${g}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{1} \\ $$

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