Previous in Relation and Functions Next in Relation and Functions
Question Number 62129 by maxmathsup by imad last updated on 15/Jun/19
$${let}\:{f}\left({x}\right)={ln}\left({x}+\mathrm{1}−\mathrm{2}\sqrt{{x}}\right) \\ $$$$\left.\mathrm{1}\right)\:{find}\:{D}_{{f}} \\ $$$$\left.\mathrm{2}\right)\:{determine}\:{f}^{−\mathrm{1}} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\:\int{f}\:\left({x}\right){dx}\:{and}\:\:\int\:{f}^{−\mathrm{1}} \left({x}\right){dx} \\ $$
Commented by arcana last updated on 16/Jun/19
$$\left.\mathrm{1}\right) \\ $$$${x}+\mathrm{1}−\mathrm{2}\sqrt{{x}}>\mathrm{0}\Rightarrow\left(\sqrt{{x}}−\mathrm{1}\right)^{\mathrm{2}} >\mathrm{0} \\ $$$${necessary}\:\sqrt{{x}}−\mathrm{1}>\mathrm{0}\:\vee\:\sqrt{{x}}−\mathrm{1}<\mathrm{0} \\ $$$$\sqrt{{x}}>\mathrm{1}\vee\sqrt{{x}}<\mathrm{1} \\ $$$${x}>\mathrm{1}\vee{x}<\mathrm{1},\:{x}\geqslant\mathrm{0} \\ $$$$\Rightarrow{D}_{{f}} =\left[\mathrm{0},\mathrm{1}\right)\cup\left(\mathrm{1},+\infty\right)=\mathbb{R}^{+} −\left\{\mathrm{1}\right\} \\ $$$$ \\ $$
Commented by kaivan.ahmadi last updated on 16/Jun/19
$${e}^{{y}} =\left(\sqrt{{x}}−\mathrm{1}\right)^{\mathrm{2}} \Rightarrow\sqrt{{e}^{{y}} }=\sqrt{{x}}−\mathrm{1}\Rightarrow\mathrm{1}+\sqrt{{e}^{{y}} }=\sqrt{{x}}\Rightarrow \\ $$$${x}=\sqrt{\mathrm{1}+\sqrt{{e}^{{y}} }}\Rightarrow{f}^{−\mathrm{1}} \left({x}\right)=\sqrt{\mathrm{1}+\sqrt{{e}^{{x}} }} \\ $$
Commented by maxmathsup by imad last updated on 16/Jun/19
![1) x∈D_f ⇔x≥0 and x−2(√x)+1>0 ⇔x≥0 and ((√x)−1)^2 >0 ⇔x≥0 and x≠1 ⇒ D_f =[0,1[∪]1,+∞[ 2)f(x)=y ⇔x =f^(−1) (y) f(x)=y ⇒ln(x+1−2(√x)) =y ⇒x+1−2(√x)=e^y ⇒((√x)−1)^(2 ) =e^y ⇒ ∣(√x)−1∣ =(√e^y ) by e^y >0 ⇒ x>1 ⇒(√x)−1 =(√e^y ) ⇒(√x)=1+e^(y/2) ⇒ x =(1+e^(y/2) )^2 ⇒ f^(−1) (x) =(1+e^(x/2) )^2 3) let I =∫ f(x)dx ⇒ I =∫ ln(x+1−2(√x))dx by parts I =xln(x+1−2(√x)) −∫ x((1−(1/(√x)))/(x+1−2(√x))) dx =xln(x+1−2(√x)) −∫ x(((√x)−1)/((√x)(x+1−2(√x)))) dx ∫ x(((√x)−1)/((√x)(x+1−2(√x))))dx = ∫ (((√x)((√x)−1))/(x+1−2(√x))) dx = ∫ (((√x)((√x)−1))/(((√x)−1)^2 )) dx =∫ ((√x)/((√x)−1)) dx =_((√x)=t) ∫ (t/(t−1))(2t)dt =2 ∫ ((t^2 −1+1)/(t−1)) dt =2 ∫(t+1)dt +2 ∫ (dt/(t−1)) =2(t^2 /2) +2t +2ln∣t−1∣ +c =t^2 +2t +2ln∣t−1∣ +c =x +2(√x) +2ln∣(√x)−1∣ +c ⇒ I =xln(x+1−2(√x))−x−2(√x)−2ln∣(√x)−1∣+c .](Q62158.png)
$$\left.\mathrm{1}\right)\:{x}\in{D}_{{f}} \:\Leftrightarrow{x}\geqslant\mathrm{0}\:{and}\:{x}−\mathrm{2}\sqrt{{x}}+\mathrm{1}>\mathrm{0}\:\Leftrightarrow{x}\geqslant\mathrm{0}\:{and}\:\left(\sqrt{{x}}−\mathrm{1}\right)^{\mathrm{2}} >\mathrm{0}\:\Leftrightarrow{x}\geqslant\mathrm{0}\:{and}\:{x}\neq\mathrm{1}\:\Rightarrow \\ $$$${D}_{{f}} =\left[\mathrm{0},\mathrm{1}\left[\cup\right]\mathrm{1},+\infty\left[\right.\right. \\ $$$$\left.\mathrm{2}\right){f}\left({x}\right)={y}\:\Leftrightarrow{x}\:={f}^{−\mathrm{1}} \left({y}\right) \\ $$$${f}\left({x}\right)={y}\:\Rightarrow{ln}\left({x}+\mathrm{1}−\mathrm{2}\sqrt{{x}}\right)\:={y}\:\Rightarrow{x}+\mathrm{1}−\mathrm{2}\sqrt{{x}}={e}^{{y}} \:\Rightarrow\left(\sqrt{{x}}−\mathrm{1}\right)^{\mathrm{2}\:} ={e}^{{y}} \:\Rightarrow \\ $$$$\mid\sqrt{{x}}−\mathrm{1}\mid\:=\sqrt{{e}^{{y}} }\:\:\:\:{by}\:{e}^{{y}} >\mathrm{0}\:\Rightarrow\:{x}>\mathrm{1}\:\Rightarrow\sqrt{{x}}−\mathrm{1}\:=\sqrt{{e}^{{y}} }\:\Rightarrow\sqrt{{x}}=\mathrm{1}+{e}^{\frac{{y}}{\mathrm{2}}} \:\Rightarrow \\ $$$${x}\:=\left(\mathrm{1}+{e}^{\frac{{y}}{\mathrm{2}}} \right)^{\mathrm{2}} \:\:\Rightarrow\:\:{f}^{−\mathrm{1}} \left({x}\right)\:=\left(\mathrm{1}+{e}^{\frac{{x}}{\mathrm{2}}} \right)^{\mathrm{2}} \\ $$$$\left.\mathrm{3}\right)\:{let}\:{I}\:=\int\:{f}\left({x}\right){dx}\:\Rightarrow\:{I}\:=\int\:\:{ln}\left({x}+\mathrm{1}−\mathrm{2}\sqrt{{x}}\right){dx}\:\:{by}\:{parts} \\ $$$${I}\:={xln}\left({x}+\mathrm{1}−\mathrm{2}\sqrt{{x}}\right)\:−\int\:{x}\frac{\mathrm{1}−\frac{\mathrm{1}}{\sqrt{{x}}}}{{x}+\mathrm{1}−\mathrm{2}\sqrt{{x}}}\:{dx} \\ $$$$={xln}\left({x}+\mathrm{1}−\mathrm{2}\sqrt{{x}}\right)\:−\int\:{x}\frac{\sqrt{{x}}−\mathrm{1}}{\sqrt{{x}}\left({x}+\mathrm{1}−\mathrm{2}\sqrt{{x}}\right)}\:{dx} \\ $$$$\int\:\:{x}\frac{\sqrt{{x}}−\mathrm{1}}{\sqrt{{x}}\left({x}+\mathrm{1}−\mathrm{2}\sqrt{{x}}\right)}{dx}\:=\:\int\:\:\frac{\sqrt{{x}}\left(\sqrt{{x}}−\mathrm{1}\right)}{{x}+\mathrm{1}−\mathrm{2}\sqrt{{x}}}\:{dx}\:=\:\int\:\frac{\sqrt{{x}}\left(\sqrt{{x}}−\mathrm{1}\right)}{\left(\sqrt{{x}}−\mathrm{1}\right)^{\mathrm{2}} }\:{dx}\:=\int\:\:\frac{\sqrt{{x}}}{\sqrt{{x}}−\mathrm{1}}\:{dx} \\ $$$$=_{\sqrt{{x}}={t}} \:\:\:\:\:\:\:\:\int\:\:\:\frac{{t}}{{t}−\mathrm{1}}\left(\mathrm{2}{t}\right){dt}\:=\mathrm{2}\:\int\:\frac{{t}^{\mathrm{2}} −\mathrm{1}+\mathrm{1}}{{t}−\mathrm{1}}\:{dt}\:=\mathrm{2}\:\int\left({t}+\mathrm{1}\right){dt}\:+\mathrm{2}\:\int\:\frac{{dt}}{{t}−\mathrm{1}} \\ $$$$=\mathrm{2}\frac{{t}^{\mathrm{2}} }{\mathrm{2}}\:+\mathrm{2}{t}\:+\mathrm{2}{ln}\mid{t}−\mathrm{1}\mid\:+{c}\:={t}^{\mathrm{2}} \:+\mathrm{2}{t}\:+\mathrm{2}{ln}\mid{t}−\mathrm{1}\mid\:+{c} \\ $$$$={x}\:+\mathrm{2}\sqrt{{x}}\:+\mathrm{2}{ln}\mid\sqrt{{x}}−\mathrm{1}\mid\:+{c}\:\Rightarrow \\ $$$${I}\:={xln}\left({x}+\mathrm{1}−\mathrm{2}\sqrt{{x}}\right)−{x}−\mathrm{2}\sqrt{{x}}−\mathrm{2}{ln}\mid\sqrt{{x}}−\mathrm{1}\mid+{c}\:. \\ $$
Commented by maxmathsup by imad last updated on 16/Jun/19
$$\int\:{f}^{−\mathrm{1}} \left({x}\right){dx}\:=\int\left(\mathrm{1}+{e}^{\frac{{x}}{\mathrm{2}}} \right)^{\mathrm{2}} {dx}\:=\int\:\:\:\left(\mathrm{1}+\mathrm{2}\:{e}^{\frac{{x}}{\mathrm{2}}} \:+{e}^{{x}} \right){dx} \\ $$$$={x}\:+\mathrm{4}{e}^{\frac{{x}}{\mathrm{2}}} \:+{e}^{{x}} \:+{c}\:. \\ $$