Question and Answers Forum

All Questions      Topic List

Mensuration Questions

Previous in All Question      Next in All Question      

Previous in Mensuration      Next in Mensuration      

Question Number 182109 by amin96 last updated on 04/Dec/22

f(x)=3x^2 −2x(√3)−8     g(x)=x^2 −(1/3)  gof^(−1) (18)=?

$$\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\mathrm{3}\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{2}\boldsymbol{{x}}\sqrt{\mathrm{3}}−\mathrm{8}\:\:\:\:\:\boldsymbol{{g}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{x}}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\boldsymbol{{gof}}^{−\mathrm{1}} \left(\mathrm{18}\right)=? \\ $$

Answered by FelipeLz last updated on 04/Dec/22

f(x) = 3x^2 −2x(√3)−8  f(x) = 3x^2 −2(√3)x+1−1−8  f(x) = ((√3)x−1)^2 −9  x = ((√3)f^(−1) (x)−1)^2 −9  f^(−1) (x) = (1/( (√3)))(1+(√(x+9)))  g○f^(−1) (x) = [(1/( (√3)))(1+(√(x+9)))]^2 −(1/3)  g○f^(−1) (x) = (1/3)(10+2(√(x+9))+x)−(1/3)  g○f^(−1) (x) = (1/3)(9+2(√(x+9))+x)  g○f^(−1) (18) = (1/3)(27+2(√(27)))  g○f^(−1) (18) = (1/3)(27+6(√3)) = 9+2(√3)

$${f}\left({x}\right)\:=\:\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}{x}\sqrt{\mathrm{3}}−\mathrm{8} \\ $$$${f}\left({x}\right)\:=\:\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{3}}{x}+\mathrm{1}−\mathrm{1}−\mathrm{8} \\ $$$${f}\left({x}\right)\:=\:\left(\sqrt{\mathrm{3}}{x}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{9} \\ $$$${x}\:=\:\left(\sqrt{\mathrm{3}}{f}^{−\mathrm{1}} \left({x}\right)−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{9} \\ $$$${f}^{−\mathrm{1}} \left({x}\right)\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left(\mathrm{1}+\sqrt{{x}+\mathrm{9}}\right) \\ $$$${g}\circ{f}^{−\mathrm{1}} \left({x}\right)\:=\:\left[\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left(\mathrm{1}+\sqrt{{x}+\mathrm{9}}\right)\right]^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${g}\circ{f}^{−\mathrm{1}} \left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{10}+\mathrm{2}\sqrt{{x}+\mathrm{9}}+{x}\right)−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$${g}\circ{f}^{−\mathrm{1}} \left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{9}+\mathrm{2}\sqrt{{x}+\mathrm{9}}+{x}\right) \\ $$$${g}\circ{f}^{−\mathrm{1}} \left(\mathrm{18}\right)\:=\:\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{27}+\mathrm{2}\sqrt{\mathrm{27}}\right) \\ $$$${g}\circ{f}^{−\mathrm{1}} \left(\mathrm{18}\right)\:=\:\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{27}+\mathrm{6}\sqrt{\mathrm{3}}\right)\:=\:\mathrm{9}+\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com