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Question Number 39702 by math khazana by abdo last updated on 10/Jul/18
let f(x)=2(√(x−(√(x−3)) +2)) 1) find D_f 2) calculate f^′ (x) 3) determine f^(−1) (x) 4) calculate (f^(−1) )^′ (x) 5) let u(x)=x^(2 ) +4 determine v(x)=fou(x) and calculate v^′ (x) 6)calculate ∫_3 ^5 f(x)dx.
$${let}\:{f}\left({x}\right)=\mathrm{2}\sqrt{{x}−\sqrt{{x}−\mathrm{3}}\:+\mathrm{2}} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{D}_{{f}} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{f}^{'} \left({x}\right) \\ $$$$\left.\mathrm{3}\right)\:{determine}\:{f}^{−\mathrm{1}} \left({x}\right) \\ $$$$\left.\mathrm{4}\right)\:{calculate}\:\left({f}^{−\mathrm{1}} \right)^{'} \left({x}\right) \\ $$$$\left.\mathrm{5}\right)\:{let}\:{u}\left({x}\right)={x}^{\mathrm{2}\:} \:+\mathrm{4} \\ $$$${determine}\:\:{v}\left({x}\right)={fou}\left({x}\right)\:{and}\:{calculate} \\ $$$${v}^{'} \left({x}\right)\: \\ $$$$\left.\mathrm{6}\right){calculate}\:\:\int_{\mathrm{3}} ^{\mathrm{5}} \:{f}\left({x}\right){dx}. \\ $$
Commented by math khazana by abdo last updated on 13/Jul/18
1) x ∈ D_f ⇔ x≥3 and x+2−(√(x−3))≥0 ⇒ x≥3 and (x+2)^2 ≥x−3 ⇒ x^2 +4x +4−x +3≥0 ⇒ x^2 +3x +7≥0 thisinequality is true because Δ= 9 −28<0 ⇒ D_f =[3,+∞[ 2) we have f^2 (x) =4(x−(√(x−3)) +2) ⇒ 2f^′ (x)f(x) =4(1 −(1/(2(√(x−3))))) =4((2(√(x−3))−1)/(2(√(x−3)))) =2((2(√(x−3)) −1)/( (√(x−3)))) ⇒f^′ (x)f(x)=((2(√(x−3))−1)/( (√(x−3)))) ⇒ f^′ (x) = ((2(√(x−3)) −1)/(2(√(x+2−(√(x−3)) )) (√(x−3)))) .
$$\left.\mathrm{1}\right)\:{x}\:\in\:{D}_{{f}} \:\Leftrightarrow\:{x}\geqslant\mathrm{3}\:{and}\:{x}+\mathrm{2}−\sqrt{{x}−\mathrm{3}}\geqslant\mathrm{0}\:\Rightarrow \\ $$$${x}\geqslant\mathrm{3}\:{and}\:\left({x}+\mathrm{2}\right)^{\mathrm{2}} \geqslant{x}−\mathrm{3}\:\Rightarrow\:{x}^{\mathrm{2}} \:+\mathrm{4}{x}\:+\mathrm{4}−{x}\:+\mathrm{3}\geqslant\mathrm{0}\:\Rightarrow \\ $$$${x}^{\mathrm{2}} \:+\mathrm{3}{x}\:+\mathrm{7}\geqslant\mathrm{0}\:{thisinequality}\:{is}\:{true}\:{because} \\ $$$$\Delta=\:\mathrm{9}\:−\mathrm{28}<\mathrm{0}\:\:\Rightarrow\:{D}_{{f}} =\left[\mathrm{3},+\infty\left[\right.\right. \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}^{\mathrm{2}} \left({x}\right)\:=\mathrm{4}\left({x}−\sqrt{{x}−\mathrm{3}}\:+\mathrm{2}\right)\:\Rightarrow \\ $$$$\mathrm{2}{f}^{'} \left({x}\right){f}\left({x}\right)\:=\mathrm{4}\left(\mathrm{1}\:−\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}−\mathrm{3}}}\right)\:=\mathrm{4}\frac{\mathrm{2}\sqrt{{x}−\mathrm{3}}−\mathrm{1}}{\mathrm{2}\sqrt{{x}−\mathrm{3}}} \\ $$$$=\mathrm{2}\frac{\mathrm{2}\sqrt{{x}−\mathrm{3}}\:−\mathrm{1}}{\:\sqrt{{x}−\mathrm{3}}}\:\Rightarrow{f}^{'} \left({x}\right){f}\left({x}\right)=\frac{\mathrm{2}\sqrt{{x}−\mathrm{3}}−\mathrm{1}}{\:\sqrt{{x}−\mathrm{3}}}\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)\:=\:\frac{\mathrm{2}\sqrt{{x}−\mathrm{3}}\:−\mathrm{1}}{\mathrm{2}\sqrt{{x}+\mathrm{2}−\sqrt{{x}−\mathrm{3}}\:}\:\sqrt{{x}−\mathrm{3}}}\:. \\ $$
Commented by math khazana by abdo last updated on 13/Jul/18
3) f^(−1) (x)=y ⇔x=f(y) ⇒x=2(√(y+2−(√(y−3)))) ⇒x^2 =4(y+2 −(√(y−3))) ⇒ x^2 =4y +8 −4(√(y−3)) ⇒ 4(√(y−3))=4y+8−x^2 ⇒ 16( y−3) = (4y+8 −x^2 )^2 ⇒ 16y −48 =(4y+8)^2 −2x^2 (4y+8)+x^4 16y −48 =16y^2 +64y +64 −8yx^(2 ) −16x^2 +x^4 ⇒ x^4 −8(y+2)x^2 +16y^2 +64−16y+48=0 ⇒ x^4 −8(y+2)x^2 +16y^2 −16y + 112 =0 Δ^′ = (−2(y+2))^2 −16y^(2 ) +16y −112 =4(y^2 +4y +4) −16 y^2 +16y −112 =−8y^2 +32y +16 −112 ...be continued...
$$\left.\mathrm{3}\right)\:{f}^{−\mathrm{1}} \left({x}\right)={y}\:\Leftrightarrow{x}={f}\left({y}\right)\:\Rightarrow{x}=\mathrm{2}\sqrt{{y}+\mathrm{2}−\sqrt{{y}−\mathrm{3}}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} \:=\mathrm{4}\left({y}+\mathrm{2}\:−\sqrt{{y}−\mathrm{3}}\right)\:\Rightarrow \\ $$$${x}^{\mathrm{2}} \:=\mathrm{4}{y}\:+\mathrm{8}\:−\mathrm{4}\sqrt{{y}−\mathrm{3}}\:\Rightarrow\:\mathrm{4}\sqrt{{y}−\mathrm{3}}=\mathrm{4}{y}+\mathrm{8}−{x}^{\mathrm{2}} \:\Rightarrow \\ $$$$\mathrm{16}\left(\:{y}−\mathrm{3}\right)\:=\:\left(\mathrm{4}{y}+\mathrm{8}\:−{x}^{\mathrm{2}} \right)^{\mathrm{2}} \:\Rightarrow \\ $$$$\mathrm{16}{y}\:−\mathrm{48}\:=\left(\mathrm{4}{y}+\mathrm{8}\right)^{\mathrm{2}} \:−\mathrm{2}{x}^{\mathrm{2}} \left(\mathrm{4}{y}+\mathrm{8}\right)+{x}^{\mathrm{4}} \\ $$$$\mathrm{16}{y}\:−\mathrm{48}\:=\mathrm{16}{y}^{\mathrm{2}} \:+\mathrm{64}{y}\:+\mathrm{64}\:−\mathrm{8}{yx}^{\mathrm{2}\:} −\mathrm{16}{x}^{\mathrm{2}} \:+{x}^{\mathrm{4}} \:\Rightarrow \\ $$$${x}^{\mathrm{4}} \:−\mathrm{8}\left({y}+\mathrm{2}\right){x}^{\mathrm{2}} \:+\mathrm{16}{y}^{\mathrm{2}} \:+\mathrm{64}−\mathrm{16}{y}+\mathrm{48}=\mathrm{0}\:\Rightarrow \\ $$$${x}^{\mathrm{4}} \:−\mathrm{8}\left({y}+\mathrm{2}\right){x}^{\mathrm{2}} \:\:+\mathrm{16}{y}^{\mathrm{2}} \:−\mathrm{16}{y}\:\:+\:\mathrm{112}\:=\mathrm{0} \\ $$$$\Delta^{'} \:=\:\left(−\mathrm{2}\left({y}+\mathrm{2}\right)\right)^{\mathrm{2}} \:−\mathrm{16}{y}^{\mathrm{2}\:} \:+\mathrm{16}{y}\:−\mathrm{112} \\ $$$$=\mathrm{4}\left({y}^{\mathrm{2}} \:+\mathrm{4}{y}\:+\mathrm{4}\right)\:−\mathrm{16}\:{y}^{\mathrm{2}} \:+\mathrm{16}{y}\:−\mathrm{112} \\ $$$$=−\mathrm{8}{y}^{\mathrm{2}} \:+\mathrm{32}{y}\:\:+\mathrm{16}\:−\mathrm{112}\:\:…{be}\:{continued}… \\ $$
Commented by maxmathsup by imad last updated on 14/Jul/18
5) we have v(x)=f(u(x)) = 2(√(u(x)−(√(u(x)−3))+2)) = 2(√(x^2 +4−(√(x^2 +1))+2)) =2(√(x^2 −(√(x^2 +1))+6)) we have v^2 (x)=4{x^2 −(√(x^2 +1))+2} ⇒ 2v(x)v^′ (x) =4{x^2 +2 −(√(x^2 +1))}^′ =4{ 2x−(x/( (√(x^2 +1))))} ⇒ v(x)v^′ (x)=2{((2x(√(x^2 +1))−x)/( (√(x^2 +1))))} ⇒v^′ (x)= (2/(v(x))) ((2x(√(x^2 +1))−x)/( (√(x^2 +1)))) = ((2x(√(x^2 +1))−x)/(((√(x^2 +6−(√(x^2 +1)))((√(x^2 +1))))))).
$$\left.\mathrm{5}\right)\:{we}\:{have}\:{v}\left({x}\right)={f}\left({u}\left({x}\right)\right)\:=\:\mathrm{2}\sqrt{{u}\left({x}\right)−\sqrt{{u}\left({x}\right)−\mathrm{3}}+\mathrm{2}} \\ $$$$=\:\mathrm{2}\sqrt{{x}^{\mathrm{2}} \:+\mathrm{4}−\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}+\mathrm{2}} \\ $$$$=\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}+\mathrm{6}} \\ $$$${we}\:{have}\:{v}^{\mathrm{2}} \left({x}\right)=\mathrm{4}\left\{{x}^{\mathrm{2}} \:−\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}+\mathrm{2}\right\}\:\Rightarrow \\ $$$$\mathrm{2}{v}\left({x}\right){v}^{'} \left({x}\right)\:=\mathrm{4}\left\{{x}^{\mathrm{2}} \:+\mathrm{2}\:−\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}\right\}^{'} =\mathrm{4}\left\{\:\mathrm{2}{x}−\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}}\right\}\:\Rightarrow \\ $$$${v}\left({x}\right){v}^{'} \left({x}\right)=\mathrm{2}\left\{\frac{\mathrm{2}{x}\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}−{x}}{\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}}\right\}\:\Rightarrow{v}^{'} \left({x}\right)=\:\frac{\mathrm{2}}{{v}\left({x}\right)}\:\frac{\mathrm{2}{x}\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}−{x}}{\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}} \\ $$$$=\:\frac{\mathrm{2}{x}\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}−{x}}{\left(\sqrt{\left.{x}^{\mathrm{2}} +\mathrm{6}−\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}\right)\left(\sqrt{\left.{x}^{\mathrm{2}} \:+\mathrm{1}\right)}\right.}\right.}. \\ $$
Commented by maxmathsup by imad last updated on 14/Jul/18
6) ∫_3 ^5 f(x)dx = 2 ∫_3 ^5 (√(x+2−(√(x−3)))) dx changement (√(x−3))=t givex=3+t^2 ∫_3 ^5 f(x)dx = 2 ∫_0 ^(√2) (√(5+t^2 −t))2tdt=4 ∫_0 ^(√2) t(√(t^2 −t +5))dt =4 ∫_0 ^(√2) t(√((t−(1/2))^2 +5−(1/4)))dt=4 ∫_0 ^(√2) t(√((t−(1/2))^2 +((19)/4)))dt =_(t−(1/2)=((√(19))/2) sh(u)) 4 ∫_(−arsh((1/( (√(19)))))) ^(argsh(((2(√2)−1)/( (√(19)))))) ((1/2) +((√(19))/2)sh(u))ch(u)((√(19))/2) ch(u)du = (√(19))∫_(−argsh((1/( (√(19)))))) ^(argsh(((2(√2)−1)/( (√(19)))))) (1+(√(19))sh(u))ch(u)du =(√(19)) ∫_(−argsh((1/( (√(19)))))) ^(argsh(((2(√2)−1)/( (√(19)))))) ch(u)du +(√(19))∫_(−argsh((1/( (√(19)))))) ^(argsh(((2(√2)−1)/( (√(19)))))) sh(u)ch(u)du =(√(19{)) ((2(√2)−1)/( (√(19)))) +(1/( (√(19))))} +((√(19))/2) { (((2(√2)−1)/( (√(19)))))^2 −((1/(19)))}.
$$\left.\mathrm{6}\right)\:\:\int_{\mathrm{3}} ^{\mathrm{5}} \:\:{f}\left({x}\right){dx}\:=\:\mathrm{2}\:\int_{\mathrm{3}} ^{\mathrm{5}} \:\sqrt{{x}+\mathrm{2}−\sqrt{{x}−\mathrm{3}}}\:{dx}\:\:{changement}\:\sqrt{{x}−\mathrm{3}}={t}\:{givex}=\mathrm{3}+{t}^{\mathrm{2}} \\ $$$$\int_{\mathrm{3}} ^{\mathrm{5}} \:\:{f}\left({x}\right){dx}\:=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \sqrt{\mathrm{5}+{t}^{\mathrm{2}} −{t}}\mathrm{2}{tdt}=\mathrm{4}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} {t}\sqrt{{t}^{\mathrm{2}} \:−{t}\:+\mathrm{5}}{dt} \\ $$$$=\mathrm{4}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} {t}\sqrt{\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\mathrm{5}−\frac{\mathrm{1}}{\mathrm{4}}}{dt}=\mathrm{4}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} {t}\sqrt{\left({t}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{19}}{\mathrm{4}}}{dt} \\ $$$$=_{{t}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{19}}}{\mathrm{2}}\:{sh}\left({u}\right)} \:\mathrm{4}\:\:\int_{−{arsh}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{19}}}\right)} ^{{argsh}\left(\frac{\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}}{\:\sqrt{\mathrm{19}}}\right)} \:\left(\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\sqrt{\mathrm{19}}}{\mathrm{2}}{sh}\left({u}\right)\right){ch}\left({u}\right)\frac{\sqrt{\mathrm{19}}}{\mathrm{2}}\:{ch}\left({u}\right){du} \\ $$$$=\:\sqrt{\mathrm{19}}\int_{−{argsh}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{19}}}\right)} ^{{argsh}\left(\frac{\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}}{\:\sqrt{\mathrm{19}}}\right)} \:\left(\mathrm{1}+\sqrt{\mathrm{19}}{sh}\left({u}\right)\right){ch}\left({u}\right){du} \\ $$$$=\sqrt{\mathrm{19}}\:\int_{−{argsh}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{19}}}\right)} ^{{argsh}\left(\frac{\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}}{\:\sqrt{\mathrm{19}}}\right)} \:{ch}\left({u}\right){du}\:\:+\sqrt{\mathrm{19}}\int_{−{argsh}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{19}}}\right)} ^{{argsh}\left(\frac{\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}}{\:\sqrt{\mathrm{19}}}\right)} \:{sh}\left({u}\right){ch}\left({u}\right){du} \\ $$$$\left.=\sqrt{\mathrm{19}\left\{\right.}\:\frac{\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}}{\:\sqrt{\mathrm{19}}}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{19}}}\right\}\:\:+\frac{\sqrt{\mathrm{19}}}{\mathrm{2}}\:\left\{\:\left(\frac{\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}}{\:\sqrt{\mathrm{19}}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{1}}{\mathrm{19}}\right)\right\}. \\ $$
Commented by ajfour last updated on 14/Jul/18
Great work sir!
$${Great}\:{work}\:{sir}! \\ $$
Commented by math khazana by abdo last updated on 15/Jul/18
thank you sir.
$${thank}\:{you}\:{sir}. \\ $$

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