Menu Close

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-




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

Leave a Reply

Your email address will not be published. Required fields are marked *