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Question Number 38721 by maxmathsup by imad last updated on 28/Jun/18
let  f(x)=(√(1+2x^2 ))  −x(√2)  +3  1) calculate lim_(x→+∞)  f(x) and lim_(x→−∞) f(x)  2)calculate lim_(x→+∞)   ((f(x))/x) and lim_(x→−∞)   ((f(x))/x)  3)give the assymtote to graph C_f   4) give the assymtote to C_f   at point A(0,f(0))  5) find f^(−1) (x) and calculate (f^(−1) )^′ (x)  6) calculate  ∫_0 ^1 f(x)dx.
$${let}\:\:{f}\left({x}\right)=\sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }\:\:−{x}\sqrt{\mathrm{2}}\:\:+\mathrm{3} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{lim}_{{x}\rightarrow+\infty} \:{f}\left({x}\right)\:{and}\:{lim}_{{x}\rightarrow−\infty} {f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right){calculate}\:{lim}_{{x}\rightarrow+\infty} \:\:\frac{{f}\left({x}\right)}{{x}}\:{and}\:{lim}_{{x}\rightarrow−\infty} \:\:\frac{{f}\left({x}\right)}{{x}} \\ $$$$\left.\mathrm{3}\right){give}\:{the}\:{assymtote}\:{to}\:{graph}\:{C}_{{f}} \\ $$$$\left.\mathrm{4}\right)\:{give}\:{the}\:{assymtote}\:{to}\:{C}_{{f}} \:\:{at}\:{point}\:{A}\left(\mathrm{0},{f}\left(\mathrm{0}\right)\right) \\ $$$$\left.\mathrm{5}\right)\:{find}\:{f}^{−\mathrm{1}} \left({x}\right)\:{and}\:{calculate}\:\left({f}^{−\mathrm{1}} \right)^{'} \left({x}\right) \\ $$$$\left.\mathrm{6}\right)\:{calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left({x}\right){dx}. \\ $$
Commented by prof Abdo imad last updated on 02/Jul/18
1) we have (√(1+2x^2  )) =x(√2)(√( 1+(1/(2x^2 ))))  ∼  x(√2){ 1+ (1/(4x^2 ))}  (x→+∞) ⇒f(x)∼ ((√2)/(4x)) +3 ⇒  lim_(x→+∞)  f(x)= 3 also we can use this method  lim_(x→+∞) f(x)=lim((((√(1+2x^2 ))−(x(√2)−3))((√(1+2x^2 )) +x(√2)−3))/( (√(1+2x^2 )) +x(√2) −3))  =lim_(x→+∞) ((1+2x^2  −(2x^2  −6x(√2) +9))/( (√(1+2x^2 )) +x(√2) −3))  =lim_(x→+∞) ((6x(√2) −8)/( (√(1+2x^2 )) +x(√2)−3))  =lim_(x→+∞)   ((6x(√2)−8)/(x(√2){(√(1+(1/(2x^2 )))) +1−(3/(x(√2)))})) =3  lim_(x→−∞) f(x)=lim_(x→−∞) (√(1+2x^2 )) −x(√2) +3 =+∞  2)we have lim_(x→+∗)  f(x)=3 ⇒lim_(x→+∞) ((f(x))/x) =0  (√(1+2x^2 ))=−x(√2)(√(1+(1/(2x^2 ))))  for x<0  ∼−x(√2)( 1+(1/(4x^2 )))(x→−∞) ⇒  f(x) ∼ −2(√2)x  −((√2)/(4x)) +3 ⇒ ((f(x))/x) ∼−2(√2) −((√2)/(4x^2 )) +(3/x)  ⇒lim_(x→+∞)   ((f(x))/x) ^ =−2(√2)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:\sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} \:}\:={x}\sqrt{\mathrm{2}}\sqrt{\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }}\:\:\sim \\ $$$${x}\sqrt{\mathrm{2}}\left\{\:\mathrm{1}+\:\frac{\mathrm{1}}{\mathrm{4}{x}^{\mathrm{2}} }\right\}\:\:\left({x}\rightarrow+\infty\right)\:\Rightarrow{f}\left({x}\right)\sim\:\frac{\sqrt{\mathrm{2}}}{\mathrm{4}{x}}\:+\mathrm{3}\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow+\infty} \:{f}\left({x}\right)=\:\mathrm{3}\:{also}\:{we}\:{can}\:{use}\:{this}\:{method} \\ $$$${lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)={lim}\frac{\left(\sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }−\left({x}\sqrt{\mathrm{2}}−\mathrm{3}\right)\right)\left(\sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }\:+{x}\sqrt{\mathrm{2}}−\mathrm{3}\right)}{\:\sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }\:+{x}\sqrt{\mathrm{2}}\:−\mathrm{3}} \\ $$$$={lim}_{{x}\rightarrow+\infty} \frac{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} \:−\left(\mathrm{2}{x}^{\mathrm{2}} \:−\mathrm{6}{x}\sqrt{\mathrm{2}}\:+\mathrm{9}\right)}{\:\sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }\:+{x}\sqrt{\mathrm{2}}\:−\mathrm{3}} \\ $$$$={lim}_{{x}\rightarrow+\infty} \frac{\mathrm{6}{x}\sqrt{\mathrm{2}}\:−\mathrm{8}}{\:\sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }\:+{x}\sqrt{\mathrm{2}}−\mathrm{3}} \\ $$$$={lim}_{{x}\rightarrow+\infty} \:\:\frac{\mathrm{6}{x}\sqrt{\mathrm{2}}−\mathrm{8}}{{x}\sqrt{\mathrm{2}}\left\{\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }}\:+\mathrm{1}−\frac{\mathrm{3}}{{x}\sqrt{\mathrm{2}}}\right\}}\:=\mathrm{3} \\ $$$${lim}_{{x}\rightarrow−\infty} {f}\left({x}\right)={lim}_{{x}\rightarrow−\infty} \sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }\:−{x}\sqrt{\mathrm{2}}\:+\mathrm{3}\:=+\infty \\ $$$$\left.\mathrm{2}\right){we}\:{have}\:{lim}_{{x}\rightarrow+\ast} \:{f}\left({x}\right)=\mathrm{3}\:\Rightarrow{lim}_{{x}\rightarrow+\infty} \frac{{f}\left({x}\right)}{{x}}\:=\mathrm{0} \\ $$$$\sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }=−{x}\sqrt{\mathrm{2}}\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} }}\:\:{for}\:{x}<\mathrm{0} \\ $$$$\sim−{x}\sqrt{\mathrm{2}}\left(\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{4}{x}^{\mathrm{2}} }\right)\left({x}\rightarrow−\infty\right)\:\Rightarrow \\ $$$${f}\left({x}\right)\:\sim\:−\mathrm{2}\sqrt{\mathrm{2}}{x}\:\:−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}{x}}\:+\mathrm{3}\:\Rightarrow\:\frac{{f}\left({x}\right)}{{x}}\:\sim−\mathrm{2}\sqrt{\mathrm{2}}\:−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}{x}^{\mathrm{2}} }\:+\frac{\mathrm{3}}{{x}} \\ $$$$\Rightarrow{lim}_{{x}\rightarrow+\infty} \:\:\frac{{f}\left({x}\right)}{{x}}\overset{} {\:}=−\mathrm{2}\sqrt{\mathrm{2}} \\ $$
Commented by prof Abdo imad last updated on 02/Jul/18
lim_(x→−∞) ((f(x))/x) =−2(√2)  but we have  f(x) ∼ −2(√2)x +3 −((√2)/(4x)) (x→−∞) ⇒  lim_(x→−∞) f(x)−(−2(√2)x+3)=0 so   y=−2(√2)+3 is equation of asshmptote to C_f   at −∞
$${lim}_{{x}\rightarrow−\infty} \frac{{f}\left({x}\right)}{{x}}\:=−\mathrm{2}\sqrt{\mathrm{2}}\:\:{but}\:{we}\:{have} \\ $$$${f}\left({x}\right)\:\sim\:−\mathrm{2}\sqrt{\mathrm{2}}{x}\:+\mathrm{3}\:−\frac{\sqrt{\mathrm{2}}}{\mathrm{4}{x}}\:\left({x}\rightarrow−\infty\right)\:\Rightarrow \\ $$$${lim}_{{x}\rightarrow−\infty} {f}\left({x}\right)−\left(−\mathrm{2}\sqrt{\mathrm{2}}{x}+\mathrm{3}\right)=\mathrm{0}\:{so}\: \\ $$$${y}=−\mathrm{2}\sqrt{\mathrm{2}}+\mathrm{3}\:{is}\:{equation}\:{of}\:{asshmptote}\:{to}\:{C}_{{f}} \\ $$$${at}\:−\infty \\ $$
Commented by prof Abdo imad last updated on 02/Jul/18
4) first change assymptote by equation of   tangent  we have f^′ (x)= ((4x)/(2(√(1+2x^2 )))) −(√2)= ((2x)/( (√(1+2x^2 )))) −(√2)  ⇒f^′ (0) =−(√2)  f(0) =4  ⇒ the equation of assymptote is  y =f^′ (0)x +f(0) =−(√2)x +4
$$\left.\mathrm{4}\right)\:{first}\:{change}\:{assymptote}\:{by}\:{equation}\:{of}\: \\ $$$${tangent} \\ $$$${we}\:{have}\:{f}^{'} \left({x}\right)=\:\frac{\mathrm{4}{x}}{\mathrm{2}\sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }}\:−\sqrt{\mathrm{2}}=\:\frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }}\:−\sqrt{\mathrm{2}} \\ $$$$\Rightarrow{f}^{'} \left(\mathrm{0}\right)\:=−\sqrt{\mathrm{2}} \\ $$$${f}\left(\mathrm{0}\right)\:=\mathrm{4}\:\:\Rightarrow\:{the}\:{equation}\:{of}\:{assymptote}\:{is} \\ $$$${y}\:={f}^{'} \left(\mathrm{0}\right){x}\:+{f}\left(\mathrm{0}\right)\:=−\sqrt{\mathrm{2}}{x}\:+\mathrm{4} \\ $$
Commented by prof Abdo imad last updated on 02/Jul/18
5) f(x)=y ⇔ x =f^(−1) (y) ⇒(√(1+2x^2 ))−x(√2)+3=y  ⇒(√(1+2x^2 ))=(x(√2)+y−3)⇒  1+2x^2  =(x(√2)+y−3)^2  ⇒  1+2x^2  =2x^2  +2x(√2)(y−3) +y^2  −6y +9 ⇒  2(√2)(y−3)x +y^(2 ) −6y +8=0 ⇒  2(√2)(y−3)x = −y^2  +6y −8 ⇒  x=((−y^2  +6y −8)/(2(√2)(y−3))) =f^(−1) (y)  ⇒  f^(−1) (x) = ((x^(2 )  −6x +8)/(2(√2)(3−x))) .  (f^(−1) (x))^′  =(1/(2(√2))){ (((2x−6)(3−x) +(x^2  −6x +8))/((3−x)^2 ))}  =(1/(2(√2))){ ((6x −2x^2  −18 +6x +x^(2 ) −6x +8)/((3−x)^2 ))}  =(1/(2(√2))){ ((−x^2   +6x  −10)/((3−x)^2 ))}
$$\left.\mathrm{5}\right)\:{f}\left({x}\right)={y}\:\Leftrightarrow\:{x}\:={f}^{−\mathrm{1}} \left({y}\right)\:\Rightarrow\sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }−{x}\sqrt{\mathrm{2}}+\mathrm{3}={y} \\ $$$$\Rightarrow\sqrt{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }=\left({x}\sqrt{\mathrm{2}}+{y}−\mathrm{3}\right)\Rightarrow \\ $$$$\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} \:=\left({x}\sqrt{\mathrm{2}}+{y}−\mathrm{3}\right)^{\mathrm{2}} \:\Rightarrow \\ $$$$\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} \:=\mathrm{2}{x}^{\mathrm{2}} \:+\mathrm{2}{x}\sqrt{\mathrm{2}}\left({y}−\mathrm{3}\right)\:+{y}^{\mathrm{2}} \:−\mathrm{6}{y}\:+\mathrm{9}\:\Rightarrow \\ $$$$\mathrm{2}\sqrt{\mathrm{2}}\left({y}−\mathrm{3}\right){x}\:+{y}^{\mathrm{2}\:} −\mathrm{6}{y}\:+\mathrm{8}=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{2}\sqrt{\mathrm{2}}\left({y}−\mathrm{3}\right){x}\:=\:−{y}^{\mathrm{2}} \:+\mathrm{6}{y}\:−\mathrm{8}\:\Rightarrow \\ $$$${x}=\frac{−{y}^{\mathrm{2}} \:+\mathrm{6}{y}\:−\mathrm{8}}{\mathrm{2}\sqrt{\mathrm{2}}\left({y}−\mathrm{3}\right)}\:={f}^{−\mathrm{1}} \left({y}\right)\:\:\Rightarrow \\ $$$${f}^{−\mathrm{1}} \left({x}\right)\:=\:\frac{{x}^{\mathrm{2}\:} \:−\mathrm{6}{x}\:+\mathrm{8}}{\mathrm{2}\sqrt{\mathrm{2}}\left(\mathrm{3}−{x}\right)}\:. \\ $$$$\left({f}^{−\mathrm{1}} \left({x}\right)\right)^{'} \:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left\{\:\frac{\left(\mathrm{2}{x}−\mathrm{6}\right)\left(\mathrm{3}−{x}\right)\:+\left({x}^{\mathrm{2}} \:−\mathrm{6}{x}\:+\mathrm{8}\right)}{\left(\mathrm{3}−{x}\right)^{\mathrm{2}} }\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left\{\:\frac{\mathrm{6}{x}\:−\mathrm{2}{x}^{\mathrm{2}} \:−\mathrm{18}\:+\mathrm{6}{x}\:+{x}^{\mathrm{2}\:} −\mathrm{6}{x}\:+\mathrm{8}}{\left(\mathrm{3}−{x}\right)^{\mathrm{2}} }\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left\{\:\frac{−{x}^{\mathrm{2}} \:\:+\mathrm{6}{x}\:\:−\mathrm{10}}{\left(\mathrm{3}−{x}\right)^{\mathrm{2}} }\right\} \\ $$$$ \\ $$

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