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Question Number 145286 by physicstutes last updated on 03/Jul/21

Given the function f(x) =((6x^2 −x^3 ))^(1/3) Find the oblique assymptote(s) of the function.

$$\mathrm{Given}\:\mathrm{the}\:\mathrm{function} \\ $$$$\:{f}\left({x}\right)\:=\sqrt[{\mathrm{3}}]{\mathrm{6}{x}^{\mathrm{2}} −{x}^{\mathrm{3}} } \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{oblique}\:\mathrm{assymptote}\left(\mathrm{s}\right)\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}. \\ $$

Answered by Olaf_Thorendsen last updated on 04/Jul/21

f(x) = ((6x^2 −x^3 ))^(1/3) f(x) = −x((1−(6/x)))^(1/3) f(x) ∼_∞ −x(1+(1/3)(−(6/x))) f(x) ∼_∞ −x+2

$${f}\left({x}\right)\:=\:\sqrt[{\mathrm{3}}]{\mathrm{6}{x}^{\mathrm{2}} −{x}^{\mathrm{3}} } \\ $$$${f}\left({x}\right)\:=\:−{x}\sqrt[{\mathrm{3}}]{\mathrm{1}−\frac{\mathrm{6}}{{x}}} \\ $$$${f}\left({x}\right)\:\underset{\infty} {\sim}\:−{x}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}\left(−\frac{\mathrm{6}}{{x}}\right)\right) \\ $$$${f}\left({x}\right)\:\underset{\infty} {\sim}\:−{x}+\mathrm{2} \\ $$

Commented by physicstutes last updated on 04/Jul/21

Thanks sir but my textbook came up with this argument which makes me confused. f(x) = ((6x^2 −x^3 ))^(1/3) f(x) = ∣x∣(((6/x)−1))^(1/3) as x →±∞, f(x) →−x hence y = −x is an oblique asymptote. How true is that sirs?

$$\mathrm{Thanks}\:\mathrm{sir}\:\mathrm{but}\:\mathrm{my}\:\mathrm{textbook}\:\mathrm{came}\:\mathrm{up}\:\mathrm{with}\:\mathrm{this}\:\mathrm{argument}\:\mathrm{which} \\ $$$$\mathrm{makes}\:\mathrm{me}\:\mathrm{confused}. \\ $$$${f}\left({x}\right)\:=\:\sqrt[{\mathrm{3}}]{\mathrm{6}{x}^{\mathrm{2}} −{x}^{\mathrm{3}} } \\ $$$${f}\left({x}\right)\:=\:\mid{x}\mid\sqrt[{\mathrm{3}}]{\frac{\mathrm{6}}{{x}}−\mathrm{1}} \\ $$$$\:\mathrm{as}\:{x}\:\rightarrow\pm\infty,\:\:{f}\left({x}\right)\:\rightarrow−{x}\:\mathrm{hence}\:{y}\:=\:−{x}\:\mathrm{is}\:\mathrm{an}\:\mathrm{oblique} \\ $$$$\mathrm{asymptote}. \\ $$$$\mathrm{How}\:\mathrm{true}\:\mathrm{is}\:\mathrm{that}\:\mathrm{sirs}? \\ $$

Answered by mathmax by abdo last updated on 04/Jul/21

lim_(x→∞) ((f(x))/x)=lim_(x→∞) ^3 (√((−x^3 +6x^2 )/x^3 ))=lim_(x→∞) (√(−1+(6/x)))=−1 lim_(x→∞) f(x)+x =lim_(x→∞) x+^3 (√(6x^2 −x^3 )) we have (−x^3 +6x^2 )^(1/3) =−x(1−(6/x))^(1/3) ∼−x(1−(2/x))=−x+2 ⇒ f(x)+x∼2 ⇒y=−x+2 is assymptote oblique for C_f

$$\mathrm{lim}_{\mathrm{x}\rightarrow\infty} \frac{\mathrm{f}\left(\mathrm{x}\right)}{\mathrm{x}}=\mathrm{lim}_{\mathrm{x}\rightarrow\infty} ^{\mathrm{3}} \sqrt{\frac{−\mathrm{x}^{\mathrm{3}} +\mathrm{6x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{3}} }}=\mathrm{lim}_{\mathrm{x}\rightarrow\infty} \sqrt{−\mathrm{1}+\frac{\mathrm{6}}{\mathrm{x}}}=−\mathrm{1} \\ $$$$\mathrm{lim}_{\mathrm{x}\rightarrow\infty} \mathrm{f}\left(\mathrm{x}\right)+\mathrm{x}\:=\mathrm{lim}_{\mathrm{x}\rightarrow\infty} \mathrm{x}+^{\mathrm{3}} \sqrt{\mathrm{6x}^{\mathrm{2}} −\mathrm{x}^{\mathrm{3}} } \\ $$$$\mathrm{we}\:\mathrm{have}\:\left(−\mathrm{x}^{\mathrm{3}} \:+\mathrm{6x}^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:=−\mathrm{x}\left(\mathrm{1}−\frac{\mathrm{6}}{\mathrm{x}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:\sim−\mathrm{x}\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{x}}\right)=−\mathrm{x}+\mathrm{2}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{x}\right)+\mathrm{x}\sim\mathrm{2}\:\Rightarrow\mathrm{y}=−\mathrm{x}+\mathrm{2}\:\mathrm{is}\:\mathrm{assymptote}\:\mathrm{oblique}\:\mathrm{for}\:\mathrm{C}_{\mathrm{f}} \\ $$

Answered by imjagoll last updated on 04/Jul/21

let y=mx+c oblique asymptotes (1) m=lim_(x→∞) (y/x) = lim_(x→∞) (((6x^2 −x^3 ))^(1/3) /x) = lim_(x→∞) (((6x^2 −x^3 ))^(1/3) /( (x^3 )^(1/3) )) = −1 (2) c = lim_(x→−∞) (y−mx) c = lim_(x→−∞) (((6x^2 −x^3 ))^(1/3) + x c = lim_(x→−∞) −x (((−6x^(−1) +1))^(1/3) −1) c = lim_(t→0) ((((6t+1))^(1/3) −1)/t) ,[− x=(1/t) ] c = lim_(t→0) ((6t)/t) × lim_(t→0) (1/( (((6t+1)^2 ))^(1/3) +1+((6t+1))^(1/3) )) c = 6 ×(1/3)= 2 thus oblique asymptotes is y =−x+2

$$\mathrm{let}\:\mathrm{y}=\mathrm{mx}+\mathrm{c}\:\mathrm{oblique}\:\mathrm{asymptotes} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{m}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{y}}{\mathrm{x}}\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{6x}^{\mathrm{2}} −\mathrm{x}^{\mathrm{3}} }}{\mathrm{x}} \\ $$$$\:=\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{6x}^{\mathrm{2}} −\mathrm{x}^{\mathrm{3}} }}{\:\sqrt[{\mathrm{3}}]{\mathrm{x}^{\mathrm{3}} }}\:=\:−\mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{c}\:=\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\left(\mathrm{y}−\mathrm{mx}\right) \\ $$$$\:\:\mathrm{c}\:=\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:\left(\sqrt[{\mathrm{3}}]{\mathrm{6x}^{\mathrm{2}} −\mathrm{x}^{\mathrm{3}} }+\:\mathrm{x}\:\right. \\ $$$$\:\mathrm{c}\:=\:\underset{{x}\rightarrow−\infty} {\mathrm{lim}}\:−\mathrm{x}\:\left(\sqrt[{\mathrm{3}}]{−\mathrm{6x}^{−\mathrm{1}} +\mathrm{1}}\:−\mathrm{1}\right) \\ $$$$\:\mathrm{c}\:=\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{6t}+\mathrm{1}}−\mathrm{1}}{\mathrm{t}}\:,\left[−\:\mathrm{x}=\frac{\mathrm{1}}{\mathrm{t}}\:\right] \\ $$$$\:\mathrm{c}\:=\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{6t}}{\mathrm{t}}\:×\:\underset{\mathrm{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\left(\mathrm{6t}+\mathrm{1}\right)^{\mathrm{2}} }+\mathrm{1}+\sqrt[{\mathrm{3}}]{\mathrm{6t}+\mathrm{1}}} \\ $$$$\:\mathrm{c}\:=\:\mathrm{6}\:×\frac{\mathrm{1}}{\mathrm{3}}=\:\mathrm{2} \\ $$$$\mathrm{thus}\:\mathrm{oblique}\:\mathrm{asymptotes}\:\mathrm{is}\: \\ $$$$\:\mathrm{y}\:=−\mathrm{x}+\mathrm{2}\: \\ $$

Commented by imjagoll last updated on 04/Jul/21

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