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Question Number 139592 by bemath last updated on 29/Apr/21

Given f(x)=(√(2+x^2 −x)) +(√(2−x^2 ))  If (g○f)(x) = 2x+1 then g^(−1) (−1)=?

$$\mathrm{Given}\:\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{2}+\mathrm{x}^{\mathrm{2}} −\mathrm{x}}\:+\sqrt{\mathrm{2}−\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{If}\:\left(\mathrm{g}\circ\mathrm{f}\right)\left(\mathrm{x}\right)\:=\:\mathrm{2x}+\mathrm{1}\:\mathrm{then}\:\mathrm{g}^{−\mathrm{1}} \left(−\mathrm{1}\right)=? \\ $$

Answered by EDWIN88 last updated on 29/Apr/21

⇒(g○f)(x)= 2x+1 ; g^(−1) (2x+1)=f(x)    { ((2x+1=−1⇒x=−1)),((g^(−1) (−1)=f(−1)=(√(2+1+1)) +(√(2−1)) = 3)) :}

$$\Rightarrow\left(\mathrm{g}\circ\mathrm{f}\right)\left(\mathrm{x}\right)=\:\mathrm{2x}+\mathrm{1}\:;\:\mathrm{g}^{−\mathrm{1}} \left(\mathrm{2x}+\mathrm{1}\right)=\mathrm{f}\left(\mathrm{x}\right) \\ $$$$\:\begin{cases}{\mathrm{2x}+\mathrm{1}=−\mathrm{1}\Rightarrow\mathrm{x}=−\mathrm{1}}\\{\mathrm{g}^{−\mathrm{1}} \left(−\mathrm{1}\right)=\mathrm{f}\left(−\mathrm{1}\right)=\sqrt{\mathrm{2}+\mathrm{1}+\mathrm{1}}\:+\sqrt{\mathrm{2}−\mathrm{1}}\:=\:\mathrm{3}}\end{cases} \\ $$

Answered by mathmax by abdo last updated on 29/Apr/21

f(x)=(√(x^2 −x+2))+(√(2−x^2 ))  gof(x)=2x+1 ⇒g(x)=f^(−1) (2x+1) ⇒g^(−1) (x)=f(2x+1) ⇒  g^(−1) (−1)=f(−1) =(√(2+1+1))+(√(2−1))=2+1=3

$$\mathrm{f}\left(\mathrm{x}\right)=\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{2}}+\sqrt{\mathrm{2}−\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{gof}\left(\mathrm{x}\right)=\mathrm{2x}+\mathrm{1}\:\Rightarrow\mathrm{g}\left(\mathrm{x}\right)=\mathrm{f}^{−\mathrm{1}} \left(\mathrm{2x}+\mathrm{1}\right)\:\Rightarrow\mathrm{g}^{−\mathrm{1}} \left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{2x}+\mathrm{1}\right)\:\Rightarrow \\ $$$$\mathrm{g}^{−\mathrm{1}} \left(−\mathrm{1}\right)=\mathrm{f}\left(−\mathrm{1}\right)\:=\sqrt{\mathrm{2}+\mathrm{1}+\mathrm{1}}+\sqrt{\mathrm{2}−\mathrm{1}}=\mathrm{2}+\mathrm{1}=\mathrm{3} \\ $$

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