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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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