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Question Number 194282 by alcohol last updated on 02/Jul/23

$$f\left(f\left(x\right)\right)=ax+b$$$$1.showthatf(ax+b)=af\left(x\right)+b$$$$deducef{}^{\prime }(ax+b)$$$$2.Showthatf{}^{\prime }\left(x\right)isaconstant$$$$hencededucef$$

Answered by Frix last updated on 02/Jul/23

$$f\left(x\right)=\alpha x+\beta \Rightarrow$$$$f\left(f\left(x\right)\right)={\alpha }^{2}x+(\alpha +1)\beta \Rightarrow$$$$\underset{[\Rightarrow a>0]}{a={\alpha }^2}\wedge b=(\alpha +1)\beta \iff \alpha =\pm \sqrt{a}\wedge \beta =\frac{b}{1\pm \sqrt{a}}\Rightarrow$$$$f\left(x\right)=\pm \sqrt{a}x+\frac{b}{1\pm \sqrt{a}}$$$$f(ax+b)=f({\alpha }^{2}x+(\alpha +1)\beta )=$$$$={\alpha }^{3}x+({\alpha }^2+\alpha +1)\beta$$$$af\left(x\right)+b={\alpha }^2(\alpha x+\beta )+(\alpha +1)\beta =$$$$={\alpha }^{3}x+({\alpha }^2+\alpha +1)\beta$$$$f^{\prime }(ax+b)={\alpha }^3=\pm a^{\frac{3}{2}}$$$$f^{\prime }\left(x\right)=\pm \sqrt{a}$$

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