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

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

Answered by Olaf_Thorendsen last updated on 04/Jul/21

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

Commented by physicstutes last updated on 04/Jul/21

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

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

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

Answered by imjagoll last updated on 04/Jul/21

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

Commented by imjagoll last updated on 04/Jul/21

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