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Question Number 150187 by mathdanisur last updated on 10/Aug/21

If f(x) = 2x^2 + 5 ⇒ f^( −1) (2) = ?

$$\mathrm{If}\:\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{2x}^{\mathrm{2}} \:+\:\mathrm{5}\:\:\Rightarrow\:\:\mathrm{f}^{\:−\mathrm{1}} \left(\mathrm{2}\right)\:=\:? \\ $$

Commented by amin96 last updated on 10/Aug/21

2x^2 +5=2 ⇒ x=±((3i)/( (√2))) f(±((3i)/( (√2))))=2 ⇒ f^(−1) (2)=±((3i)/( (√2)))

$$\mathrm{2}{x}^{\mathrm{2}} +\mathrm{5}=\mathrm{2}\:\:\Rightarrow\:\:\:{x}=\pm\frac{\mathrm{3}{i}}{\:\sqrt{\mathrm{2}}}\:\:\: \\ $$$${f}\left(\pm\frac{\mathrm{3}{i}}{\:\sqrt{\mathrm{2}}}\right)=\mathrm{2}\:\:\:\:\Rightarrow\:\:{f}^{−\mathrm{1}} \left(\mathrm{2}\right)=\pm\frac{\mathrm{3}{i}}{\:\sqrt{\mathrm{2}}} \\ $$

Answered by puissant last updated on 10/Aug/21

f(x)=y ⇒ 2x^2 +5 = y ⇒ 2x^2 = y−5 ⇒ x=±(√((1/2)(y−5))) f^(−1) (x)=±((√2)/2)(√(x−5))...

$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{y} \\ $$$$\Rightarrow\:\mathrm{2x}^{\mathrm{2}} +\mathrm{5}\:=\:\mathrm{y} \\ $$$$\Rightarrow\:\mathrm{2x}^{\mathrm{2}} =\:\mathrm{y}−\mathrm{5} \\ $$$$\Rightarrow\:\mathrm{x}=\pm\sqrt{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{y}−\mathrm{5}\right)} \\ $$$$\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right)=\pm\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\sqrt{\mathrm{x}−\mathrm{5}}... \\ $$

Commented by puissant last updated on 10/Aug/21

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Commented by Rasheed.Sindhi last updated on 10/Aug/21

Perhaps there′s a typo sir x=±(√((1/2)(y−5)))⇒f^(−1) (x)=±((√(2(x−5)))/2) f^(−1) (2) will not be real.

$${Perhaps}\:{there}'{s}\:{a}\:{typo}\:{sir} \\ $$$$\mathrm{x}=\pm\sqrt{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{y}−\mathrm{5}\right)}\Rightarrow\mathrm{f}^{−\mathrm{1}} \left(\mathrm{x}\right)=\pm\frac{\sqrt{\mathrm{2}\left(\mathrm{x}−\mathrm{5}\right)}}{\mathrm{2}} \\ $$$$\mathrm{f}^{−\mathrm{1}} \left(\mathrm{2}\right)\:\mathrm{will}\:\mathrm{not}\:\mathrm{be}\:\mathrm{real}. \\ $$

Commented by Rasheed.Sindhi last updated on 10/Aug/21

you′re welcome sir

$$\mathrm{you}'{re}\:\mathrm{welcome}\:\mathrm{sir} \\ $$

Commented by mathdanisur last updated on 10/Aug/21

thankyou Ser

$$\mathrm{thankyou}\:\mathrm{Ser} \\ $$

Answered by liberty last updated on 10/Aug/21

let f^(−1) (2)=t ⇒f(t)=2 ⇒2t^2 +5=2 ⇒t^2 =−(3/2) ; t= ± i(√(3/2))

$$\mathrm{let}\:\mathrm{f}^{−\mathrm{1}} \left(\mathrm{2}\right)=\mathrm{t}\:\Rightarrow\mathrm{f}\left(\mathrm{t}\right)=\mathrm{2} \\ $$$$\Rightarrow\mathrm{2t}^{\mathrm{2}} +\mathrm{5}=\mathrm{2} \\ $$$$\Rightarrow\mathrm{t}^{\mathrm{2}} =−\frac{\mathrm{3}}{\mathrm{2}}\:;\:\mathrm{t}=\:\pm\:{i}\sqrt{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$

Commented by mathdanisur last updated on 10/Aug/21

thankyou Ser

$${thankyou}\:{Ser} \\ $$

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