Question Number 42631 by maxmathsup by imad last updated on 29/Aug/18
$${let}\:{f}\left({x}\right)=\mathrm{2}\sqrt{{x}−\mathrm{1}}\:−\mathrm{2}{x} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{D}_{{f}} \\ $$$$\left.\mathrm{2}\right)\:{study}\:{the}\:{variation}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{3}\:\right)\:{calculate}\:\:\int_{\mathrm{1}} ^{\mathrm{3}} \:{f}\left({x}\right){dx} \\ $$$$\left.\mathrm{4}\right)\:{determine}\:{f}^{−\mathrm{1}} \left({x}\right)\:{and}\:{calculate}\:\:\:\int_{\mathrm{1}} ^{\mathrm{3}} \:{f}^{−\mathrm{1}} \left({x}\right){dx} \\ $$$$\left.\mathrm{5}\right)\:\:{find}\:{the}\:{values}\:{of}\:\:{A}\:=\:\:\int_{\mathrm{1}} ^{\mathrm{3}} \:\:\:\frac{{f}\left({x}\right)}{{f}^{−\mathrm{1}} \left({x}\right){dx}}\:{and}\: \\ $$$${B}\:=\:\frac{\int_{\mathrm{1}} ^{\mathrm{3}} \:\:{f}\left({x}\right)}{\int_{\mathrm{1}} ^{\mathrm{3}} \:{f}^{−\mathrm{1}} \left({x}\right)}\:{dx}. \\ $$
Commented by bbnnftm833 last updated on 30/Aug/18
$${we}\:{need}\:{the}\:{solution} \\ $$
Commented by maxmathsup by imad last updated on 31/Aug/18
$$\left.\mathrm{1}\right)\:{x}\in\:{D}_{{f}} \:\Leftrightarrow\:{x}−\mathrm{1}\geqslant\mathrm{0}\:\Rightarrow{D}_{{f}} =\left[\mathrm{1},+\infty\left[\right.\right. \\ $$$$\left.\mathrm{2}\right){lim}_{{x}\rightarrow+\infty} \:{f}\left({x}\right)\:\:={lim}_{{x}\rightarrow+\infty} \:\mathrm{2}{x}\left(\:\sqrt{\frac{{x}−\mathrm{1}}{{x}^{\mathrm{2}} }}\:−\mathrm{1}\right)={lim}_{{x}\rightarrow+\infty} \mathrm{2}{x}\left(\sqrt{\frac{\mathrm{1}}{{x}}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}−\mathrm{1}\right) \\ $$$$={lim}_{{x}\rightarrow+\infty} \:−\mathrm{2}{x}\:=−\infty \\ $$$${we}\:{have}\:{f}^{'} \left({x}\right)=\:\frac{\mathrm{1}}{\:\sqrt{{x}−\mathrm{1}}}\:−\mathrm{2}\:{for}\:{x}>\mathrm{1}\:\Rightarrow{f}^{'} \left({x}\right)=\frac{\mathrm{1}−\mathrm{2}\sqrt{{x}−\mathrm{1}}}{\:\sqrt{{x}−\mathrm{1}}} \\ $$$${f}^{'} \left({x}\right)=\mathrm{0}\:\Leftrightarrow\:\mathrm{2}\sqrt{{x}−\mathrm{1}}=\mathrm{1}\:\Leftrightarrow\sqrt{{x}−\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}\:\Leftrightarrow{x}−\mathrm{1}=\frac{\mathrm{1}}{\mathrm{4}}\:\Leftrightarrow{x}\:=\frac{\mathrm{5}}{\mathrm{4}} \\ $$$${f}^{'} \left({x}\right)>\mathrm{0}\:\Leftrightarrow\:\mathrm{1}−\mathrm{2}\sqrt{{x}−\mathrm{1}}>\mathrm{0}\:\Leftrightarrow\mathrm{2}\sqrt{{x}−\mathrm{1}}<\mathrm{1}\:\Leftrightarrow\:\mathrm{1}<{x}<\frac{\mathrm{5}}{\mathrm{4}} \\ $$$${f}^{'} \left({x}\right)<\mathrm{0}\:\Leftrightarrow\:{x}>\frac{\mathrm{5}}{\mathrm{4}} \\ $$$${x}_{} \:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{5}}{\mathrm{4}}\:\:\:\:\:\:\:\:\:\:\:\:+\infty \\ $$$${f}^{'} \left({x}\right)\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:− \\ $$$${f}\left({x}\right)\:\:\:\:−\mathrm{2}\:\:\:\:\:{inc}\:\:−\frac{\mathrm{3}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{decr}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{f}\left(\mathrm{1}\right)=−\mathrm{2}\:\:\:\:\:{f}\left(\frac{\mathrm{5}}{\mathrm{4}}\right)=\mathrm{2}\sqrt{\frac{\mathrm{5}}{\mathrm{4}}−\mathrm{1}}−\mathrm{2}\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$=\mathrm{1}−\frac{\mathrm{5}}{\mathrm{2}}\:=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 31/Aug/18
![3) ∫_1 ^3 f(x)dx = ∫_1 ^3 {2(√(x−1))−2x}dx = 2 [(2/3)(x−1)^(3/2) ]_1 ^3 − [x^2 ]_1 ^3 =(4/3){ 2^(3/2) } −(9−1) = (4/3)(√2^3 ) −8 =(4/3)(2(√2)) −8 =((8(√2))/3) −8 .](https://www.tinkutara.com/question/Q42662.png)
$$\left.\mathrm{3}\right)\:\int_{\mathrm{1}} ^{\mathrm{3}} \:{f}\left({x}\right){dx}\:=\:\int_{\mathrm{1}} ^{\mathrm{3}} \:\left\{\mathrm{2}\sqrt{{x}−\mathrm{1}}−\mathrm{2}{x}\right\}{dx}\:=\:\mathrm{2}\:\left[\frac{\mathrm{2}}{\mathrm{3}}\left({x}−\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right]_{\mathrm{1}} ^{\mathrm{3}} −\:\:\left[{x}^{\mathrm{2}} \right]_{\mathrm{1}} ^{\mathrm{3}} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}\left\{\:\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{2}}} \right\}\:−\left(\mathrm{9}−\mathrm{1}\right)\:=\:\frac{\mathrm{4}}{\mathrm{3}}\sqrt{\mathrm{2}^{\mathrm{3}} }\:\:−\mathrm{8}\:\:=\frac{\mathrm{4}}{\mathrm{3}}\left(\mathrm{2}\sqrt{\mathrm{2}}\right)\:−\mathrm{8}\:=\frac{\mathrm{8}\sqrt{\mathrm{2}}}{\mathrm{3}}\:−\mathrm{8}\:. \\ $$
Commented by maxmathsup by imad last updated on 31/Aug/18
$$\left.\mathrm{4}\right)\:\:{f}\left({x}\right)={y}\:\Leftrightarrow\:{x}\:={f}^{−\mathrm{1}} \left({y}\right)\:\:\:{and}\:{f}\left({x}\right)={y}\:\Rightarrow\mathrm{2}\sqrt{{x}−\mathrm{1}}−\mathrm{2}{x}={y}\:\:\:\left({x}\geqslant\mathrm{1}\right)\:\Rightarrow \\ $$$$\mathrm{2}\sqrt{{x}−\mathrm{1}}\:=\mathrm{2}{x}+{y}\:\Rightarrow\:\mathrm{4}\left({x}−\mathrm{1}\right)\:=\left(\mathrm{2}{x}+{y}\right)^{\mathrm{2}} \:\Rightarrow\mathrm{4}{x}−\mathrm{4}\:=\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{4}{xy}\:+{y}^{\mathrm{2}} \:\Rightarrow \\ $$$$\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{4}{xy}\:+{y}^{\mathrm{2}} \:−\mathrm{4}{x}+\mathrm{4}\:=\mathrm{0}\:\Rightarrow\:\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{4}\left({y}−\mathrm{1}\right){x}\:+{y}^{\mathrm{2}} \:+\mathrm{4}\:=\mathrm{0} \\ $$$$\Delta^{'} \:=\mathrm{4}\left({y}−\mathrm{1}\right)^{\mathrm{2}} \:−\mathrm{4}\left({y}^{\mathrm{2}} \:+\mathrm{4}\right)\:=\mathrm{4}{y}^{\mathrm{2}} \:−\mathrm{8}{y}\:+\mathrm{4}\:−\mathrm{4}{y}^{\mathrm{2}} \:−\mathrm{16}\:=−\mathrm{8}{y}\:−\mathrm{12} \\ $$$$\:{this}\:{equation}\:{have}\:{solution}\:\Leftrightarrow−\mathrm{8}{y}−\mathrm{12}\geqslant\mathrm{0}\:\Leftrightarrow\mathrm{8}{y}\:+\mathrm{12}\leqslant\mathrm{0}\:\Leftrightarrow{y}\:\leqslant\frac{−\mathrm{12}}{\mathrm{8}}\left(=−\frac{\mathrm{3}}{\mathrm{2}}\right) \\ $$$${x}_{\mathrm{1}} =\frac{−\mathrm{2}\left({y}−\mathrm{1}\right)+\sqrt{−\mathrm{8}{y}−\mathrm{12}}}{\mathrm{4}}\:=\frac{−\left({y}−\mathrm{1}\right)\:+\sqrt{−\mathrm{2}{y}−\mathrm{3}}}{\mathrm{2}} \\ $$$${x}_{\mathrm{2}} =\frac{−\left({y}−\mathrm{1}\right)−\sqrt{−\mathrm{2}{y}−\mathrm{3}}}{\mathrm{2}}\: \\ $$$${x}_{\mathrm{1}} −\mathrm{1}\:=\frac{\mathrm{1}−{y}\:+\sqrt{−\mathrm{2}{y}−\mathrm{3}}}{\mathrm{2}}\:−\mathrm{1}\:=\frac{−\mathrm{1}−{y}\:+\sqrt{−\mathrm{2}{y}\:−\mathrm{3}}}{\mathrm{2}} \\ $$$${but}\:\:{y}\leqslant−\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow−{y}\geqslant\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow−\mathrm{1}−{y}\:\geqslant\frac{\mathrm{1}}{\mathrm{2}}\:\geqslant\mathrm{0}\:\Rightarrow{x}_{\mathrm{1}} −\mathrm{1}\geqslant\mathrm{0}\:\:{no}\:{need}\:{to}\:{verify}\:{x}_{\mathrm{2}} \\ $$$$\Rightarrow{f}^{−\mathrm{1}} \left({y}\right)\:=\frac{\mathrm{1}−{y}\:+\sqrt{−\mathrm{2}{y}−\mathrm{3}}}{\mathrm{2}}\:\:\Rightarrow{f}^{−\mathrm{1}} \left({x}\right)\:=\frac{\mathrm{1}−{x}+\sqrt{−\mathrm{2}{x}−\mathrm{3}}}{\mathrm{2}} \\ $$$${with}\:{x}\leqslant−\frac{\mathrm{3}}{\mathrm{2}}\:. \\ $$$$ \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 31/Aug/18
$$\left.\mathrm{4}\right)\:{forgive}\:\:{the}\:{Q}\:{is}\:{calculate}\:\int_{−\mathrm{3}} ^{−\mathrm{2}} \:{f}^{−\mathrm{1}} \left({x}\right){dx} \\ $$$$\left.\mathrm{5}\right)\:{the}\:{Q}\:{is}\:{find}\:{the}\:{value}\:{of}\:\:{A}\:=\:\frac{\int_{\mathrm{2}} ^{\mathrm{3}} \:{f}\left({x}\right){dx}}{\int_{−\mathrm{3}} ^{−\mathrm{2}} \:{f}^{−\mathrm{1}} \left({x}\right){dx}} \\ $$$${B}\:=\:\:\int\:\:\:\:\:\frac{{f}\left({x}\right)}{{f}^{−\mathrm{1}} \left({x}\right)}{dx}\:. \\ $$
Commented by maxmathsup by imad last updated on 31/Aug/18
![4) we have f^(−1) (x)=((1−x+(√(−2x−3)))/2) ⇒ ∫_(−3) ^(−2) f^(−1) (x)dx =(1/2) ∫_(−3) ^(−2) (1−x) +(1/2) ∫_(−3) ^(−2) (√(−2x−3))dx but ∫_(−3) ^(−2) (1−x)dx =[x−(x^2 /2)]_(−3) ^(−2) =−2−2−(−3−(9/2))=−1 +(9/2) =(7/2) and ∫_(−3) ^(−2) (√(−2x−3))dx =[−(1/3)(−2x−3)^(3/2) ]_(−3) ^(−2) =−(1/3){1−(3)^(3/2) } =(√3) −(1/3) ⇒∫_(−3) ^(−2) f^(−1) (x)dx = (7/4) +((√3)/2) −(1/6) =((38)/(24)) +((√3)/2) =((19)/(12)) +((√3)/2)](https://www.tinkutara.com/question/Q42675.png)
$$\left.\mathrm{4}\right)\:{we}\:{have}\:{f}^{−\mathrm{1}} \left({x}\right)=\frac{\mathrm{1}−{x}+\sqrt{−\mathrm{2}{x}−\mathrm{3}}}{\mathrm{2}}\:\Rightarrow \\ $$$$\int_{−\mathrm{3}} ^{−\mathrm{2}} \:\:{f}^{−\mathrm{1}} \left({x}\right){dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\mathrm{3}} ^{−\mathrm{2}} \left(\mathrm{1}−{x}\right)\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{−\mathrm{3}} ^{−\mathrm{2}} \:\sqrt{−\mathrm{2}{x}−\mathrm{3}}{dx}\:\:{but}\:\: \\ $$$$\int_{−\mathrm{3}} ^{−\mathrm{2}} \left(\mathrm{1}−{x}\right){dx}\:=\left[{x}−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right]_{−\mathrm{3}} ^{−\mathrm{2}} \:\:=−\mathrm{2}−\mathrm{2}−\left(−\mathrm{3}−\frac{\mathrm{9}}{\mathrm{2}}\right)=−\mathrm{1}\:+\frac{\mathrm{9}}{\mathrm{2}}\:=\frac{\mathrm{7}}{\mathrm{2}}\:\:\:{and} \\ $$$$\int_{−\mathrm{3}} ^{−\mathrm{2}} \:\sqrt{−\mathrm{2}{x}−\mathrm{3}}{dx}\:=\left[−\frac{\mathrm{1}}{\mathrm{3}}\left(−\mathrm{2}{x}−\mathrm{3}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right]_{−\mathrm{3}} ^{−\mathrm{2}} \:=−\frac{\mathrm{1}}{\mathrm{3}}\left\{\mathrm{1}−\left(\mathrm{3}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right\} \\ $$$$=\sqrt{\mathrm{3}}\:−\frac{\mathrm{1}}{\mathrm{3}}\:\Rightarrow\int_{−\mathrm{3}} ^{−\mathrm{2}} \:{f}^{−\mathrm{1}} \left({x}\right){dx}\:=\:\frac{\mathrm{7}}{\mathrm{4}}\:\:+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{6}}\:=\frac{\mathrm{38}}{\mathrm{24}}\:+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:=\frac{\mathrm{19}}{\mathrm{12}}\:+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 31/Aug/18
![5) we have ∫_2 ^3 f(x)dx = 2 ∫_2 ^3 (√(x−1)) −∫_2 ^3 2x dx =2[(2/3)(x−1)^(3/2) ]_2 ^3 −[x^2 ]_2 ^3 =(4/3){ 2^(3/2) −1}−{9−4} =(4/3)(2(√2)) −5 =((8(√2))/3) −5 also we have ∫_(−3) ^(−2) f^(−1) (x)dx =((19)/(12)) +((√3)/2) ⇒ A = ((∫_2 ^3 f(x)dx)/(∫_(−3) ^(−2) f^(−1) (x)dx)) = ((((8(√2))/3)−5)/(((19)/(12)) +((√3)/2))) .](https://www.tinkutara.com/question/Q42676.png)
$$\left.\mathrm{5}\right)\:{we}\:{have}\:\:\int_{\mathrm{2}} ^{\mathrm{3}} \:{f}\left({x}\right){dx}\:=\:\mathrm{2}\:\int_{\mathrm{2}} ^{\mathrm{3}} \:\sqrt{{x}−\mathrm{1}}\:−\int_{\mathrm{2}} ^{\mathrm{3}} \:\mathrm{2}{x}\:{dx}\: \\ $$$$=\mathrm{2}\left[\frac{\mathrm{2}}{\mathrm{3}}\left({x}−\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right]_{\mathrm{2}} ^{\mathrm{3}} \:−\left[{x}^{\mathrm{2}} \right]_{\mathrm{2}} ^{\mathrm{3}} \:\:\:=\frac{\mathrm{4}}{\mathrm{3}}\left\{\:\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{2}}} \:−\mathrm{1}\right\}−\left\{\mathrm{9}−\mathrm{4}\right\} \\ $$$$=\frac{\mathrm{4}}{\mathrm{3}}\left(\mathrm{2}\sqrt{\mathrm{2}}\right)\:−\mathrm{5}\:=\frac{\mathrm{8}\sqrt{\mathrm{2}}}{\mathrm{3}}\:−\mathrm{5}\:\:{also}\:{we}\:{have}\:\:\int_{−\mathrm{3}} ^{−\mathrm{2}} \:\:{f}^{−\mathrm{1}} \left({x}\right){dx}\:=\frac{\mathrm{19}}{\mathrm{12}}\:+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\Rightarrow \\ $$$${A}\:=\:\frac{\int_{\mathrm{2}} ^{\mathrm{3}} \:{f}\left({x}\right){dx}}{\int_{−\mathrm{3}} ^{−\mathrm{2}} \:{f}^{−\mathrm{1}} \left({x}\right){dx}}\:=\:\frac{\frac{\mathrm{8}\sqrt{\mathrm{2}}}{\mathrm{3}}−\mathrm{5}}{\frac{\mathrm{19}}{\mathrm{12}}\:+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\:. \\ $$
Answered by behi83417@gmail.com last updated on 30/Aug/18
![1)x−1≥0⇒x≥1⇒D_f =[1,+∞) 2)y=2(√(x−1))−2x y′=(1/( (√(x−1))))−2⇒x≠1⇒D_f =(1,+∞) x−1=t^2 ⇒x=t^2 +1 f(t)=2t−2(t^2 +1)=−2(t^2 −t+1) ⇒t^2 −t+1=((−y)/2)⇒(t−(1/2))^2 =((−y)/2)−(3/4) ⇒t−(1/2)=±(√(−((2y+3)/4)))⇒t=(1/2)(1±(√(−2y−3))) ⇒(√(x−1))=(1/2)(1±(√(−2y−3))) ⇒x=1+(1/4)(1±(√(−2y−3)))^2 ⇒f^(−1) (x)=1+(1/4)(1±(√(−2x−3)))^2 −2y−3≥0⇒2y+3≤0⇒y≤−(3/2) ⇒R_f =(−∞,−(3/2)]. 3)∫_1 ^3 f(x)dx=∫_1 ^3 (2(√(x−1))−2x)dx= =[(4/3)(x−1)^(3/2) −x^2 ]_1 ^3 =8(((√2)/3)−1). 4)∫f^(−1) (x)dx=∫[1+(1/4)(1±(√(−2x−3)))^2 ]dx= =∫[1+(1/4)(1∓(2x+3)±2(√(−2x−3)))]dx I_1 =∫[1+(1/4)(2x+4+2(√(−2x−3)))]dx= =(1/2)(2x+(((x+1)^2 )/2)−(1/3)(−2x−3)^(3/2) )_1 ^3 = =(1/2)[6+8−27i−2−2+((i(√5))/3)]= =5−((81−(√5))/6).i](https://www.tinkutara.com/question/Q42660.png)
$$\left.\mathrm{1}\right){x}−\mathrm{1}\geqslant\mathrm{0}\Rightarrow{x}\geqslant\mathrm{1}\Rightarrow{D}_{{f}} =\left[\mathrm{1},+\infty\right) \\ $$$$\left.\mathrm{2}\right){y}=\mathrm{2}\sqrt{{x}−\mathrm{1}}−\mathrm{2}{x} \\ $$$${y}'=\frac{\mathrm{1}}{\:\sqrt{{x}−\mathrm{1}}}−\mathrm{2}\Rightarrow{x}\neq\mathrm{1}\Rightarrow{D}_{{f}} =\left(\mathrm{1},+\infty\right) \\ $$$${x}−\mathrm{1}={t}^{\mathrm{2}} \Rightarrow{x}={t}^{\mathrm{2}} +\mathrm{1} \\ $$$${f}\left({t}\right)=\mathrm{2}{t}−\mathrm{2}\left({t}^{\mathrm{2}} +\mathrm{1}\right)=−\mathrm{2}\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right) \\ $$$$\Rightarrow{t}^{\mathrm{2}} −{t}+\mathrm{1}=\frac{−{y}}{\mathrm{2}}\Rightarrow\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} =\frac{−{y}}{\mathrm{2}}−\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow{t}−\frac{\mathrm{1}}{\mathrm{2}}=\pm\sqrt{−\frac{\mathrm{2}{y}+\mathrm{3}}{\mathrm{4}}}\Rightarrow{t}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}\pm\sqrt{−\mathrm{2}{y}−\mathrm{3}}\right) \\ $$$$\Rightarrow\sqrt{{x}−\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}\pm\sqrt{−\mathrm{2}{y}−\mathrm{3}}\right) \\ $$$$\Rightarrow{x}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}\pm\sqrt{−\mathrm{2}{y}−\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{f}^{−\mathrm{1}} \left({x}\right)=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}\pm\sqrt{−\mathrm{2}{x}−\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$−\mathrm{2}{y}−\mathrm{3}\geqslant\mathrm{0}\Rightarrow\mathrm{2}{y}+\mathrm{3}\leqslant\mathrm{0}\Rightarrow{y}\leqslant−\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow{R}_{{f}} =\left(−\infty,−\frac{\mathrm{3}}{\mathrm{2}}\right]. \\ $$$$\left.\mathrm{3}\right)\underset{\mathrm{1}} {\overset{\mathrm{3}} {\int}}{f}\left({x}\right){dx}=\underset{\mathrm{1}} {\overset{\mathrm{3}} {\int}}\left(\mathrm{2}\sqrt{{x}−\mathrm{1}}−\mathrm{2}{x}\right){dx}= \\ $$$$=\left[\frac{\mathrm{4}}{\mathrm{3}}\left({x}−\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} −{x}^{\mathrm{2}} \underset{\mathrm{1}} {\overset{\mathrm{3}} {\right]}}=\mathrm{8}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{3}}−\mathrm{1}\right). \\ $$$$\left.\mathrm{4}\right)\int{f}^{−\mathrm{1}} \left({x}\right){dx}=\int\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}\pm\sqrt{−\mathrm{2}{x}−\mathrm{3}}\right)^{\mathrm{2}} \right]{dx}= \\ $$$$=\int\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}\mp\left(\mathrm{2}{x}+\mathrm{3}\right)\pm\mathrm{2}\sqrt{−\mathrm{2}{x}−\mathrm{3}}\right)\right]{dx} \\ $$$${I}_{\mathrm{1}} =\int\left[\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}{x}+\mathrm{4}+\mathrm{2}\sqrt{−\mathrm{2}{x}−\mathrm{3}}\right)\right]{dx}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{x}+\frac{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}\left(−\mathrm{2}{x}−\mathrm{3}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \underset{\mathrm{1}} {\overset{\mathrm{3}} {\right)}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{6}+\mathrm{8}−\mathrm{27}{i}−\mathrm{2}−\mathrm{2}+\frac{{i}\sqrt{\mathrm{5}}}{\mathrm{3}}\right]= \\ $$$$=\mathrm{5}−\frac{\mathrm{81}−\sqrt{\mathrm{5}}}{\mathrm{6}}.{i}\: \\ $$