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

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

$$forf\left(g\left(x\right)\right)={(x^2+1)}^{50}thereisno$$$$uniquesolutionforf\left(x\right)andf\left(x\right).$$

Answered by SdC355 last updated on 13/Mar/25

$$$$$$(f\circ \mathrm{g})\left(x\right)={(x^2+1)}^{50}$$$$\mathrm{we}\mathrm{can}\mathrm{renote}(f\circ \mathrm{g})\left(x\right)\mathrm{as}f\left(\mathrm{g}\left(x\right)\right)$$$$f\left(\mathrm{g}\left(x\right)\right)={(x^2+1)}^{50}$$$$\mathrm{but}\mathrm{we}{\mathrm{can}}^{\prime }\mathrm{t}\mathrm{find}f\left(x\right)\mathrm{and}\mathrm{g}\left(x\right).....$$$$\mathrm{because}\mathrm{it}{\mathrm{dosen}}^{\prime }\mathrm{t}\mathrm{matter}\mathrm{if}$$$$f\left(x\right)=x^{50}\mathrm{and}\mathrm{g}\left(x\right)=x^2+1\mathrm{and}$$$$f\left(x\right)=x^{25}\mathrm{and}\mathrm{g}\left(x\right)={(x^2+1)}^2$$$$\mathrm{So}\mathrm{we}{\mathrm{can}}^{\prime }\mathrm{t}\mathrm{Find}\mathrm{a}\mathrm{specific}\mathrm{function}$$$$f\mathrm{and}\mathrm{g}$$

Answered by MathematicalUser2357 last updated on 13/Mar/25

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

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