Question Number 151460 by mathdanisur last updated on 21/Aug/21
![if f(x) = x^2 - 2x + 2 and g(x) = (√x) + 1 find [ f o g ] (x) = ?](https://www.tinkutara.com/question/Q151460.png)
$$\mathrm{if}\:\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{x}^{\mathrm{2}} \:-\:\mathrm{2x}\:+\:\mathrm{2} \\ $$$$\mathrm{and}\:\:\mathrm{g}\left(\mathrm{x}\right)\:=\:\sqrt{\mathrm{x}}\:+\:\mathrm{1} \\ $$$$\mathrm{find}\:\:\:\left[\:\mathrm{f}\:{o}\:\mathrm{g}\:\right]\:\left(\mathrm{x}\right)\:=\:? \\ $$
Answered by puissant last updated on 21/Aug/21
=f(g(x)) =((√x)+1)^2 −2((√x)+1)+2 = x+2(√x)+1−2(√x)−2+2 [fog](x) = x+1..](https://www.tinkutara.com/question/Q151463.png)
$${f}\left({x}\right)={x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}\:;\:{g}\left({x}\right)=\sqrt{{x}}+\mathrm{1} \\ $$$$\left[{fog}\right]\left({x}\right)={f}\left({g}\left({x}\right)\right) \\ $$$$=\left(\sqrt{{x}}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}\left(\sqrt{{x}}+\mathrm{1}\right)+\mathrm{2} \\ $$$$=\:{x}+\mathrm{2}\sqrt{{x}}+\mathrm{1}−\mathrm{2}\sqrt{{x}}−\mathrm{2}+\mathrm{2} \\ $$$$\left[{fog}\right]\left({x}\right)\:=\:{x}+\mathrm{1}.. \\ $$
Commented by mathdanisur last updated on 21/Aug/21
$$\mathrm{Thank}\:\mathrm{you}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$