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Question Number 154200 by amin96 last updated on 15/Sep/21
g(((x−1)/(x+1)))=((7x+3)/(x+1)) and f(x^2 −2x+3)=3x^2 −6x+7 find (f+g)(x)=?
$${g}\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)=\frac{\mathrm{7}{x}+\mathrm{3}}{{x}+\mathrm{1}}\:\:{and}\:\:{f}\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}\right)=\mathrm{3}{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{7} \\ $$$${find}\:\:\left({f}+{g}\right)\left({x}\right)=?\:\:\: \\ $$
Answered by Rasheed.Sindhi last updated on 15/Sep/21
g(((x−1)/(x+1)))=((7x+3)/(x+1)) and f(x^2 −2x+3)=3x^2 −6x+7 find (f+g)(x)=? ▶ ((x−1)/(x+1))=y⇒x−1=xy+y⇒x=((y+1)/(1−y)) g(y)=((7(((y+1)/(1−y)))+3)/((((y+1)/(1−y)))+1))=((7y+7+3−3y)/(y+1+1−y))=((4y+10)/2) g(y)=2y+5⇒g(x)=2x+5 ▶f(x^2 −2x+3)=3x^2 −6x+7=3x^2 −6x+9−2 f(x^2 −2x+3)=3(x^2 −2x+3)−2 Let x^2 −2x+3=y f(y)=3y−2⇒f(x)=3x−2 (f+g)(x)=(2x+5)+(3x−2) ▶(f+g)(x)=5x+3
$${g}\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)=\frac{\mathrm{7}{x}+\mathrm{3}}{{x}+\mathrm{1}}\:\:{and}\:\:{f}\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}\right)=\mathrm{3}{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{7} \\ $$$${find}\:\:\left({f}+{g}\right)\left({x}\right)=?\:\:\: \\ $$$$\blacktriangleright\:\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}={y}\Rightarrow{x}−\mathrm{1}={xy}+{y}\Rightarrow{x}=\frac{{y}+\mathrm{1}}{\mathrm{1}−{y}} \\ $$$${g}\left({y}\right)=\frac{\mathrm{7}\left(\frac{{y}+\mathrm{1}}{\mathrm{1}−{y}}\right)+\mathrm{3}}{\left(\frac{{y}+\mathrm{1}}{\mathrm{1}−{y}}\right)+\mathrm{1}}=\frac{\mathrm{7}{y}+\mathrm{7}+\mathrm{3}−\mathrm{3}{y}}{{y}+\mathrm{1}+\mathrm{1}−{y}}=\frac{\mathrm{4}{y}+\mathrm{10}}{\mathrm{2}} \\ $$$${g}\left({y}\right)=\mathrm{2}{y}+\mathrm{5}\Rightarrow{g}\left({x}\right)=\mathrm{2}{x}+\mathrm{5} \\ $$$$\blacktriangleright{f}\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}\right)=\mathrm{3}{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{7}=\mathrm{3}{x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{9}−\mathrm{2} \\ $$$${f}\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}\right)=\mathrm{3}\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}\right)−\mathrm{2} \\ $$$${Let}\:{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}={y} \\ $$$${f}\left({y}\right)=\mathrm{3}{y}−\mathrm{2}\Rightarrow{f}\left({x}\right)=\mathrm{3}{x}−\mathrm{2} \\ $$$$\left({f}+{g}\right)\left({x}\right)=\left(\mathrm{2}{x}+\mathrm{5}\right)+\left(\mathrm{3}{x}−\mathrm{2}\right) \\ $$$$\blacktriangleright\left({f}+{g}\right)\left({x}\right)=\mathrm{5}{x}+\mathrm{3} \\ $$
Commented by amin96 last updated on 15/Sep/21
nice work. thanks sir
$${nice}\:{work}.\:{thanks}\:{sir} \\ $$
Commented by Rasheed.Sindhi last updated on 15/Sep/21
You′re welcome!
$$\mathrm{You}'\mathrm{re}\:\mathrm{welcome}! \\ $$

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