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Question Number 46419 by maxmathsup by imad last updated on 25/Oct/18
let f(x)= (√(x+1−(√(x−1)))) 1) find D_f 2) determine f^(−1) (x) 3) find ∫ f(x)dx 4) dtetrmine ∫ f^(−1) (x) 5) let g(x)= (ch(x))^2 calculate fog(x) and (fog)^′ (x) .
$${let}\:{f}\left({x}\right)=\:\sqrt{{x}+\mathrm{1}−\sqrt{{x}−\mathrm{1}}} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{D}_{{f}} \\ $$$$\left.\mathrm{2}\right)\:{determine}\:{f}^{−\mathrm{1}} \left({x}\right) \\ $$$$\left.\mathrm{3}\right)\:{find}\:\int\:{f}\left({x}\right){dx} \\ $$$$\left.\mathrm{4}\right)\:{dtetrmine}\:\int\:\:{f}^{−\mathrm{1}} \left({x}\right) \\ $$$$\left.\mathrm{5}\right)\:{let}\:{g}\left({x}\right)=\:\left({ch}\left({x}\right)\right)^{\mathrm{2}} \:\:\:{calculate}\:{fog}\left({x}\right)\:{and}\:\left({fog}\right)^{'} \left({x}\right)\:. \\ $$
Commented by maxmathsup by imad last updated on 28/Oct/18
1) x∈ D_f ⇔x+1−(√(x−1))≥0 and x≥1 ⇒x+1≥(√(x−1)) and x≥1 ⇒(x+1)^2 ≥x−1 and x≥1 ⇒x^2 +2x+1−x+1≥0 and x≥1 ⇒x(x+1)≥0 and x≥1 ⇒ D_f =[1,+∞[ 2)let f(x)=y ⇒x=f^(−1) (y) f(x)=y ⇒x+1−(√(x−1))=y^2 ⇒(x+1)−y^2 =(√(x−1))⇒ (x+1−y^2 )^2 =x−1 ⇒(x+1)^2 −2y^2 (x+1)+y^4 −x+1=0 ⇒ x^2 +2x+1 −2y^2 x−2y^2 +y^4 −x+1=0 ⇒ x^2 +(1−2y^2 )x +y^4 −2y^2 +2=0 this equation have s^t ⇔Δ≥0 ⇒ (1−2y^2 )^2 −4(y^4 −2y^2 +2)≥0 ⇒4y^4 −4y^2 +1−4y^4 +8y^2 −8 ≥0 ⇒ 4y^2 −7 ≥0 so x_1 =((2y^2 −1+(√(4y^2 −7)))/2) and x_2 =((2y^2 −1−(√(4y^2 −7)))/2) f^(−1) (x)=((2x^2 −1 +^− (√(4x^2 −7)))/2)
$$\left.\mathrm{1}\right)\:{x}\in\:{D}_{{f}} \:\Leftrightarrow{x}+\mathrm{1}−\sqrt{{x}−\mathrm{1}}\geqslant\mathrm{0}\:{and}\:{x}\geqslant\mathrm{1}\:\Rightarrow{x}+\mathrm{1}\geqslant\sqrt{{x}−\mathrm{1}}\:{and}\:{x}\geqslant\mathrm{1} \\ $$$$\Rightarrow\left({x}+\mathrm{1}\right)^{\mathrm{2}} \geqslant{x}−\mathrm{1}\:{and}\:{x}\geqslant\mathrm{1}\:\Rightarrow{x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{1}−{x}+\mathrm{1}\geqslant\mathrm{0}\:{and}\:{x}\geqslant\mathrm{1}\: \\ $$$$\Rightarrow{x}\left({x}+\mathrm{1}\right)\geqslant\mathrm{0}\:{and}\:{x}\geqslant\mathrm{1}\:\Rightarrow\:{D}_{{f}} =\left[\mathrm{1},+\infty\left[\right.\right. \\ $$$$\left.\mathrm{2}\right){let}\:{f}\left({x}\right)={y}\:\Rightarrow{x}={f}^{−\mathrm{1}} \left({y}\right) \\ $$$${f}\left({x}\right)={y}\:\Rightarrow{x}+\mathrm{1}−\sqrt{{x}−\mathrm{1}}={y}^{\mathrm{2}} \:\Rightarrow\left({x}+\mathrm{1}\right)−{y}^{\mathrm{2}} =\sqrt{{x}−\mathrm{1}}\Rightarrow \\ $$$$\left({x}+\mathrm{1}−{y}^{\mathrm{2}} \right)^{\mathrm{2}} ={x}−\mathrm{1}\:\Rightarrow\left({x}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}{y}^{\mathrm{2}} \left({x}+\mathrm{1}\right)+{y}^{\mathrm{4}} −{x}+\mathrm{1}=\mathrm{0}\:\Rightarrow \\ $$$${x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{1}\:−\mathrm{2}{y}^{\mathrm{2}} {x}−\mathrm{2}{y}^{\mathrm{2}} \:+{y}^{\mathrm{4}} −{x}+\mathrm{1}=\mathrm{0}\:\Rightarrow \\ $$$${x}^{\mathrm{2}} \:+\left(\mathrm{1}−\mathrm{2}{y}^{\mathrm{2}} \right){x}\:+{y}^{\mathrm{4}} −\mathrm{2}{y}^{\mathrm{2}} \:+\mathrm{2}=\mathrm{0}\:{this}\:{equation}\:{have}\:{s}^{{t}} \:\:\:\Leftrightarrow\Delta\geqslant\mathrm{0}\:\Rightarrow \\ $$$$\left(\mathrm{1}−\mathrm{2}{y}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}\left({y}^{\mathrm{4}} −\mathrm{2}{y}^{\mathrm{2}} +\mathrm{2}\right)\geqslant\mathrm{0}\:\Rightarrow\mathrm{4}{y}^{\mathrm{4}} −\mathrm{4}{y}^{\mathrm{2}} +\mathrm{1}−\mathrm{4}{y}^{\mathrm{4}} \:+\mathrm{8}{y}^{\mathrm{2}} −\mathrm{8}\:\geqslant\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{4}{y}^{\mathrm{2}} −\mathrm{7}\:\geqslant\mathrm{0}\:\:{so}\:{x}_{\mathrm{1}} =\frac{\mathrm{2}{y}^{\mathrm{2}} −\mathrm{1}+\sqrt{\mathrm{4}{y}^{\mathrm{2}} −\mathrm{7}}}{\mathrm{2}}\:\:{and}\:{x}_{\mathrm{2}} =\frac{\mathrm{2}{y}^{\mathrm{2}} −\mathrm{1}−\sqrt{\mathrm{4}{y}^{\mathrm{2}} −\mathrm{7}}}{\mathrm{2}} \\ $$$${f}^{−\mathrm{1}} \left({x}\right)=\frac{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\:\overset{−} {+}\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{7}}}{\mathrm{2}} \\ $$
Commented by maxmathsup by imad last updated on 28/Oct/18
3) ∫ f(x)dx = ∫ (√(x−1+2−(√(x−1))))dx changement (√(x−1))=t givex−1=t^2 ⇒ ∫ f(x)dx = ∫ (√(t^2 +2−t))2t dt =2 ∫t (√(t^2 −t +2))dt =2 ∫ t(√(t^2 −2(1/2)t +(1/4)+2−(1/4)))dt = 2 ∫ t(√((t−(1/2))^2 +(7/4)))dt =_(t−(1/2)=((√7)/2)sh(t)) 2 ∫ (((√7)/2)sh(t)+(1/2))((√7)/2)ch(t)((√7)/2)ch(t)dt =(7/4) ∫ (1+(√7)sh(t)ch^2 (t) dt =(7/4) ∫ ch^2 t dt +((7(√7))/4) ∫ sh(t)ch^2 (t)dt but ∫ ch^2 t dt =∫ ((1+ch(2t))/2)dt =(t/2) +(1/4)sh(2t)=(t/2) +(1/2)sh(t)ch(t) and ∫ sh(t)ch^2 t dt =(1/3)ch^3 t ⇒ ∫ f(x)dx =(7/8)t +(7/8)sht)ch(t) +((7(√7))/(12))ch^3 (t) +c =(7/8)(√(x−1)) +(7/8)sh((√(x−1)))ch((√(x−1)))+((7(√7))/(12))ch^3 ((√(x−1))) +c .
$$\left.\mathrm{3}\right)\:\int\:{f}\left({x}\right){dx}\:=\:\int\:\sqrt{{x}−\mathrm{1}+\mathrm{2}−\sqrt{{x}−\mathrm{1}}}{dx}\:\:{changement}\:\sqrt{{x}−\mathrm{1}}={t}\:{givex}−\mathrm{1}={t}^{\mathrm{2}} \Rightarrow \\ $$$$\int\:{f}\left({x}\right){dx}\:=\:\int\:\sqrt{{t}^{\mathrm{2}} \:+\mathrm{2}−{t}}\mathrm{2}{t}\:{dt}\:=\mathrm{2}\:\int{t}\:\sqrt{{t}^{\mathrm{2}} −{t}\:+\mathrm{2}}{dt} \\ $$$$=\mathrm{2}\:\int\:\:{t}\sqrt{{t}^{\mathrm{2}} −\mathrm{2}\frac{\mathrm{1}}{\mathrm{2}}{t}\:+\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{2}−\frac{\mathrm{1}}{\mathrm{4}}}{dt}\:=\:\mathrm{2}\:\int\:{t}\sqrt{\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{7}}{\mathrm{4}}}{dt} \\ $$$$=_{{t}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{7}}}{\mathrm{2}}{sh}\left({t}\right)} \:\:\:\mathrm{2}\:\:\int\:\left(\frac{\sqrt{\mathrm{7}}}{\mathrm{2}}{sh}\left({t}\right)+\frac{\mathrm{1}}{\mathrm{2}}\right)\frac{\sqrt{\mathrm{7}}}{\mathrm{2}}{ch}\left({t}\right)\frac{\sqrt{\mathrm{7}}}{\mathrm{2}}{ch}\left({t}\right){dt} \\ $$$$=\frac{\mathrm{7}}{\mathrm{4}}\:\int\:\left(\mathrm{1}+\sqrt{\mathrm{7}}{sh}\left({t}\right){ch}^{\mathrm{2}} \left({t}\right)\:{dt}\right. \\ $$$$=\frac{\mathrm{7}}{\mathrm{4}}\:\int\:{ch}^{\mathrm{2}} {t}\:{dt}\:\:+\frac{\mathrm{7}\sqrt{\mathrm{7}}}{\mathrm{4}}\:\int\:{sh}\left({t}\right){ch}^{\mathrm{2}} \left({t}\right){dt}\:\:{but}\:\int\:{ch}^{\mathrm{2}} {t}\:{dt}\:=\int\:\frac{\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)}{\mathrm{2}}{dt} \\ $$$$=\frac{{t}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}{sh}\left(\mathrm{2}{t}\right)=\frac{{t}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}{sh}\left({t}\right){ch}\left({t}\right)\:{and}\:\:\int\:\:{sh}\left({t}\right){ch}^{\mathrm{2}} {t}\:{dt}\:=\frac{\mathrm{1}}{\mathrm{3}}{ch}^{\mathrm{3}} {t}\:\:\Rightarrow \\ $$$$\left.\int\:{f}\left({x}\right){dx}\:=\frac{\mathrm{7}}{\mathrm{8}}{t}\:+\frac{\mathrm{7}}{\mathrm{8}}{sht}\right){ch}\left({t}\right)\:+\frac{\mathrm{7}\sqrt{\mathrm{7}}}{\mathrm{12}}{ch}^{\mathrm{3}} \left({t}\right)\:+{c} \\ $$$$=\frac{\mathrm{7}}{\mathrm{8}}\sqrt{{x}−\mathrm{1}}\:+\frac{\mathrm{7}}{\mathrm{8}}{sh}\left(\sqrt{\left.{x}−\mathrm{1}\right)}{ch}\left(\sqrt{{x}−\mathrm{1}}\right)+\frac{\mathrm{7}\sqrt{\mathrm{7}}}{\mathrm{12}}{ch}^{\mathrm{3}} \left(\sqrt{{x}−\mathrm{1}}\right)\:+{c}\:.\right. \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 28/Oct/18
4) we have 2∫ f^(−1) (x)dx= ∫ (2x^2 −1)dx +^− ∫ (√(4x^2 −7))dx but ∫ (2x^2 −1)dx =(2/3)x^3 −x +c_1 ∫ (√(4x^2 −7))dx=∫2(√(x^2 −(7/4)))dx =_(x=((√7)/2)ch(t)) 2 ∫ ((√7)/2)sh(t)((√7)/2)sh(t)dt =(7/2) ∫ sh^2 (t)dt =(7/4) ∫ (ch(2t)−1)dt =(7/8)sh(2t)−((7t)/4) +c_2 =(7/4)sh(t)ch(t)−(7/4)t +c_2 =(7/4) (√((((2x)/( (√7) )))^2 −1))((2x)/( (√7))) −(7/4) argch(((2x)/( (√7))))+c_2 =((√7)/2)(√(((4x^2 )/7)−1))−(7/4)ln(((2x)/( (√7))) +(√(((4x^2 )/7)−1))) +c_2 ⇒ ∫ f^(−1) (x)dx =(x^3 /3) −(x/2) +^− {((√7)/4)(√(((4x^2 )/7)−1)) −(7/8)ln(((2x)/( (√7))) +(√(((4x^2 )/7)−1))) +C .
$$\left.\mathrm{4}\right)\:{we}\:{have}\:\mathrm{2}\int\:\:{f}^{−\mathrm{1}} \left({x}\right){dx}=\:\int\:\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\right){dx}\:\overset{−} {+}\:\int\:\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{7}}{dx}\:{but} \\ $$$$\int\:\left(\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\right){dx}\:=\frac{\mathrm{2}}{\mathrm{3}}{x}^{\mathrm{3}} −{x}\:+{c}_{\mathrm{1}} \\ $$$$\int\:\sqrt{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{7}}{dx}=\int\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\frac{\mathrm{7}}{\mathrm{4}}}{dx}\:=_{{x}=\frac{\sqrt{\mathrm{7}}}{\mathrm{2}}{ch}\left({t}\right)} \:\:\mathrm{2}\:\int\:\:\frac{\sqrt{\mathrm{7}}}{\mathrm{2}}{sh}\left({t}\right)\frac{\sqrt{\mathrm{7}}}{\mathrm{2}}{sh}\left({t}\right){dt} \\ $$$$=\frac{\mathrm{7}}{\mathrm{2}}\:\int\:\:{sh}^{\mathrm{2}} \left({t}\right){dt}\:=\frac{\mathrm{7}}{\mathrm{4}}\:\int\:\left({ch}\left(\mathrm{2}{t}\right)−\mathrm{1}\right){dt} \\ $$$$=\frac{\mathrm{7}}{\mathrm{8}}{sh}\left(\mathrm{2}{t}\right)−\frac{\mathrm{7}{t}}{\mathrm{4}}\:+{c}_{\mathrm{2}} \\ $$$$=\frac{\mathrm{7}}{\mathrm{4}}{sh}\left({t}\right){ch}\left({t}\right)−\frac{\mathrm{7}}{\mathrm{4}}{t}\:+{c}_{\mathrm{2}} =\frac{\mathrm{7}}{\mathrm{4}}\:\sqrt{\left(\frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{7}}\:}\right)^{\mathrm{2}} −\mathrm{1}}\frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{7}}}\:−\frac{\mathrm{7}}{\mathrm{4}}\:{argch}\left(\frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{7}}}\right)+{c}_{\mathrm{2}} \\ $$$$=\frac{\sqrt{\mathrm{7}}}{\mathrm{2}}\sqrt{\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{7}}−\mathrm{1}}−\frac{\mathrm{7}}{\mathrm{4}}{ln}\left(\frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{7}}}\:+\sqrt{\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{7}}−\mathrm{1}}\right)\:+{c}_{\mathrm{2}} \:\:\Rightarrow \\ $$$$\int\:{f}^{−\mathrm{1}} \left({x}\right){dx}\:=\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\:−\frac{{x}}{\mathrm{2}}\:\overset{−} {+}\:\left\{\frac{\sqrt{\mathrm{7}}}{\mathrm{4}}\sqrt{\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{7}}−\mathrm{1}}\:−\frac{\mathrm{7}}{\mathrm{8}}{ln}\left(\frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{7}}}\:+\sqrt{\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{7}}−\mathrm{1}}\right)\:+{C}\:.\right. \\ $$
Commented by maxmathsup by imad last updated on 28/Oct/18
error of typo ∫ f^(−1) (x)dx=(x^3 /3) −(x/2) +^− {((√7)/4) x(√(((4x^2 )/7)−1)) −(7/8)ln(((2x)/( (√7))) +(√(((4x^2 )/7)−1))) +C
$${error}\:{of}\:{typo}\: \\ $$$$\int\:{f}^{−\mathrm{1}} \left({x}\right){dx}=\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\:−\frac{{x}}{\mathrm{2}}\:\overset{−} {+}\left\{\frac{\sqrt{\mathrm{7}}}{\mathrm{4}}\:{x}\sqrt{\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{7}}−\mathrm{1}}\:−\frac{\mathrm{7}}{\mathrm{8}}{ln}\left(\frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{7}}}\:+\sqrt{\frac{\mathrm{4}{x}^{\mathrm{2}} }{\mathrm{7}}−\mathrm{1}}\right)\:+{C}\:\right. \\ $$

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