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Question Number 37892 by abdo mathsup 649 cc last updated on 19/Jun/18

let f(x)=(√(x+(√(x+1)))) 1) find D_f 2) give the equation of assymtote at point A(0,f(o)) 3) if f(x)∼ a(x−1) +b (x→1) determine a andb 4) calculate f^′ (x) 5) find f^(−1) (x) and (f^(−1) )^′ (x)

$${let}\:\:\:{f}\left({x}\right)=\sqrt{{x}+\sqrt{{x}+\mathrm{1}}} \\ $$$$\left.\mathrm{1}\right)\:{find}\:\:{D}_{{f}} \\ $$$$\left.\mathrm{2}\right)\:\:{give}\:{the}\:{equation}\:{of}\:{assymtote}\:{at}\:{point} \\ $$$${A}\left(\mathrm{0},{f}\left({o}\right)\right) \\ $$$$\left.\mathrm{3}\right)\:{if}\:{f}\left({x}\right)\sim\:{a}\left({x}−\mathrm{1}\right)\:\:+{b}\:\:\left({x}\rightarrow\mathrm{1}\right)\:{determine}\:{a}\:{andb} \\ $$$$\left.\mathrm{4}\right)\:{calculate}\:{f}^{'} \left({x}\right) \\ $$$$\left.\mathrm{5}\right)\:{find}\:\:{f}^{−\mathrm{1}} \left({x}\right)\:\:{and}\:\:\left({f}^{−\mathrm{1}} \right)^{'} \left({x}\right) \\ $$

Commented by prof Abdo imad last updated on 21/Jun/18

4)we have f^2 (x)=x +(√(x+1)) ⇒ 2f(x)f^′ (x)= 1 +(1/(2(√(x+1)))) ⇒ 2(√(x+(√(x+1)))) f^′ (x)= 1+(1/(2(√(x+1)))) ⇒ f^′ (x)= (1/(2(√(x+(√(x+1)))))){ 1+(1/(2(√(x+1))))} 5) let f(x)=y ⇔ x=f^(−1) (y) ⇒ (√(x+(√(x+1))))=y ⇒ x+(√(x+1))=y^2 ⇒ (√(x+1))=y^2 −x ⇒x+1=(y^2 −x)^2 =x+1 ⇒ x^2 −2y^2 x +y^4 −x−1=0 ⇒x^2 −(2y^2 +1)x +y^4 −1=0 Δ=(2y^2 +1)^2 −4(y^4 −1) =4y^4 +4y^2 +1−4y^4 +4=4y^2 +5 >0 x_1 =((2y^2 +1 +(√(4y^2 +5)))/2) x_2 = ((2y^2 +1 −(√(4y^2 +5)))/2) but we must have x+1≥0 and y^2 −x ≥0 after verification we find that x=((2y^2 +1−(√(4y^2 +5)))/2) ⇒ f^(−1) (x)=((2x^2 +1 −(√(4x^2 +5)))/2) so (f^(−1) )^′ (x)= 2x −(1/2) ((8x)/(2(√(4x^2 +5)))) =2x −((4x)/(√(4x^2 +5))) .

$$\left.\mathrm{4}\right){we}\:{have}\:{f}^{\mathrm{2}} \left({x}\right)={x}\:+\sqrt{{x}+\mathrm{1}}\:\Rightarrow \\ $$$$\mathrm{2}{f}\left({x}\right){f}^{'} \left({x}\right)=\:\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}+\mathrm{1}}}\:\Rightarrow \\ $$$$\mathrm{2}\sqrt{{x}+\sqrt{{x}+\mathrm{1}}}\:{f}^{'} \left({x}\right)=\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}+\mathrm{1}}}\:\Rightarrow\: \\ $$$${f}^{'} \left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}+\sqrt{{x}+\mathrm{1}}}}\left\{\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}+\mathrm{1}}}\right\} \\ $$$$\left.\mathrm{5}\right)\:{let}\:{f}\left({x}\right)={y}\:\:\Leftrightarrow\:{x}={f}^{−\mathrm{1}} \left({y}\right)\:\Rightarrow \\ $$$$\sqrt{{x}+\sqrt{{x}+\mathrm{1}}}={y}\:\:\Rightarrow\:{x}+\sqrt{{x}+\mathrm{1}}={y}^{\mathrm{2}} \:\Rightarrow \\ $$$$\sqrt{{x}+\mathrm{1}}={y}^{\mathrm{2}} −{x}\:\Rightarrow{x}+\mathrm{1}=\left({y}^{\mathrm{2}} −{x}\right)^{\mathrm{2}} ={x}+\mathrm{1}\:\Rightarrow \\ $$$${x}^{\mathrm{2}} \:−\mathrm{2}{y}^{\mathrm{2}} {x}\:+{y}^{\mathrm{4}} \:−{x}−\mathrm{1}=\mathrm{0}\:\Rightarrow{x}^{\mathrm{2}} \:−\left(\mathrm{2}{y}^{\mathrm{2}} +\mathrm{1}\right){x}\:+{y}^{\mathrm{4}} −\mathrm{1}=\mathrm{0} \\ $$$$\Delta=\left(\mathrm{2}{y}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} \:−\mathrm{4}\left({y}^{\mathrm{4}} −\mathrm{1}\right) \\ $$$$=\mathrm{4}{y}^{\mathrm{4}} \:+\mathrm{4}{y}^{\mathrm{2}} \:+\mathrm{1}−\mathrm{4}{y}^{\mathrm{4}} \:+\mathrm{4}=\mathrm{4}{y}^{\mathrm{2}} \:+\mathrm{5}\:>\mathrm{0} \\ $$$${x}_{\mathrm{1}} =\frac{\mathrm{2}{y}^{\mathrm{2}} \:+\mathrm{1}\:+\sqrt{\mathrm{4}{y}^{\mathrm{2}} \:+\mathrm{5}}}{\mathrm{2}} \\ $$$${x}_{\mathrm{2}} =\:\frac{\mathrm{2}{y}^{\mathrm{2}} \:+\mathrm{1}\:−\sqrt{\mathrm{4}{y}^{\mathrm{2}} \:+\mathrm{5}}}{\mathrm{2}}\:\:{but}\:{we}\:{must}\:{have} \\ $$$${x}+\mathrm{1}\geqslant\mathrm{0}\:\:{and}\:{y}^{\mathrm{2}} −{x}\:\geqslant\mathrm{0}\:{after}\:{verification} \\ $$$${we}\:{find}\:{that}\:{x}=\frac{\mathrm{2}{y}^{\mathrm{2}} \:+\mathrm{1}−\sqrt{\mathrm{4}{y}^{\mathrm{2}} \:+\mathrm{5}}}{\mathrm{2}}\:\Rightarrow \\ $$$${f}^{−\mathrm{1}} \left({x}\right)=\frac{\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{1}\:−\sqrt{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{5}}}{\mathrm{2}}\:\:{so} \\ $$$$\left({f}^{−\mathrm{1}} \right)^{'} \left({x}\right)=\:\mathrm{2}{x}\:−\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\mathrm{8}{x}}{\mathrm{2}\sqrt{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{5}}} \\ $$$$=\mathrm{2}{x}\:−\frac{\mathrm{4}{x}}{\sqrt{\mathrm{4}{x}^{\mathrm{2}} \:+\mathrm{5}}}\:. \\ $$$$ \\ $$

Commented by math khazana by abdo last updated on 21/Jun/18

1) x ∈ D_f ⇔ x+1≥0 and x+(√(x+1))≥0 ⇒ x ≥−1 and x +(√(x+1))≥0 so if x ≥0 we get x+(√(x+1)) >0 if −1≤x≤0 (√(x+1)) +x =(√(x+1)) −(−x) and ((√(x+1)))^2 −(−x)^2 =x+1 −x^2 =−x^2 +x +1 ⇒Δ=1−4(−1)=5 x_1 =((−1+(√5))/2) andx_2 = ((−1−(√5))/2) (√(x+1)) +x = ((x+1 −x^2 )/((√(x+1))−x)) and (√(x+1)) −x>0 so (√(1+x))+x ≥0 ⇔ x ∈[((−1−(√5))/2) , ((−1+(√5))/2)] but [−1,0]⊂[((−1−(√5))/2),((−1+(√5))/2)] ⇒D_f =[−1,+∞[ 2) y =f^′ (0)x +f(0) but f(x)=(√(x+(√(x+1)))) ⇒ f(0)=1 we have f^2 (x)=x +(√(x+1)) ⇒ 2f(x)f^′ (x)=1+ (1/(2(√(x+1)))) ⇒2f(0)f^′ (0)= (3/2) ⇒ f^′ (0) =(3/4) so the equation of assymptote is y= (3/4) x+1 .

$$\left.\mathrm{1}\right)\:{x}\:\in\:{D}_{{f}} \:\Leftrightarrow\:{x}+\mathrm{1}\geqslant\mathrm{0}\:{and}\:{x}+\sqrt{{x}+\mathrm{1}}\geqslant\mathrm{0}\:\Rightarrow \\ $$$${x}\:\geqslant−\mathrm{1}\:{and}\:{x}\:+\sqrt{{x}+\mathrm{1}}\geqslant\mathrm{0}\:\:{so}\:{if}\:{x}\:\geqslant\mathrm{0}\:{we}\:{get} \\ $$$${x}+\sqrt{{x}+\mathrm{1}}\:>\mathrm{0}\:\:{if}\:−\mathrm{1}\leqslant{x}\leqslant\mathrm{0} \\ $$$$\sqrt{{x}+\mathrm{1}}\:\:+{x}\:=\sqrt{{x}+\mathrm{1}}\:−\left(−{x}\right)\:{and} \\ $$$$\left(\sqrt{{x}+\mathrm{1}}\right)^{\mathrm{2}} \:−\left(−{x}\right)^{\mathrm{2}} ={x}+\mathrm{1}\:−{x}^{\mathrm{2}} \\ $$$$=−{x}^{\mathrm{2}} \:+{x}\:+\mathrm{1}\:\Rightarrow\Delta=\mathrm{1}−\mathrm{4}\left(−\mathrm{1}\right)=\mathrm{5} \\ $$$${x}_{\mathrm{1}} =\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:{andx}_{\mathrm{2}} =\:\frac{−\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\sqrt{{x}+\mathrm{1}}\:+{x}\:=\:\frac{{x}+\mathrm{1}\:−{x}^{\mathrm{2}} }{\sqrt{{x}+\mathrm{1}}−{x}}\:\:\:{and}\:\sqrt{{x}+\mathrm{1}}\:−{x}>\mathrm{0}\:{so} \\ $$$$\sqrt{\mathrm{1}+{x}}+{x}\:\geqslant\mathrm{0}\:\Leftrightarrow\:{x}\:\in\left[\frac{−\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\:,\:\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right]\:{but} \\ $$$$\left[−\mathrm{1},\mathrm{0}\right]\subset\left[\frac{−\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}},\frac{−\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right]\:\Rightarrow{D}_{{f}} =\left[−\mathrm{1},+\infty\left[\right.\right. \\ $$$$\left.\mathrm{2}\right)\:{y}\:={f}^{'} \left(\mathrm{0}\right){x}\:+{f}\left(\mathrm{0}\right)\:\:{but}\:\:{f}\left({x}\right)=\sqrt{{x}+\sqrt{{x}+\mathrm{1}}}\:\Rightarrow \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{1}\:{we}\:{have}\:{f}^{\mathrm{2}} \left({x}\right)={x}\:+\sqrt{{x}+\mathrm{1}}\:\Rightarrow \\ $$$$\mathrm{2}{f}\left({x}\right){f}^{'} \left({x}\right)=\mathrm{1}+\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}+\mathrm{1}}}\:\Rightarrow\mathrm{2}{f}\left(\mathrm{0}\right){f}^{'} \left(\mathrm{0}\right)=\:\frac{\mathrm{3}}{\mathrm{2}}\:\Rightarrow \\ $$$${f}^{'} \left(\mathrm{0}\right)\:=\frac{\mathrm{3}}{\mathrm{4}}\:{so}\:{the}\:{equation}\:{of}\:{assymptote}\:{is} \\ $$$${y}=\:\frac{\mathrm{3}}{\mathrm{4}}\:{x}+\mathrm{1}\:. \\ $$

Commented by math khazana by abdo last updated on 21/Jun/18

3) we have a=f^′ (1) and b=f(1) b=(√(1+(√2))) 2f(1)f^′ (1)=1+ (1/(2(√2))) ⇒ f^′ (1)=(1/(2(√(1+(√2)))))(1+(1/(2(√2)))) =a

$$\left.\mathrm{3}\right)\:{we}\:{have}\:{a}={f}^{'} \left(\mathrm{1}\right)\:{and}\:{b}={f}\left(\mathrm{1}\right) \\ $$$${b}=\sqrt{\mathrm{1}+\sqrt{\mathrm{2}}} \\ $$$$\mathrm{2}{f}\left(\mathrm{1}\right){f}^{'} \left(\mathrm{1}\right)=\mathrm{1}+\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\Rightarrow\:{f}^{'} \left(\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{1}+\sqrt{\mathrm{2}}}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\right)\:={a} \\ $$

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