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Question Number 194837 by mr W last updated on 16/Jul/23

$$forx>0findtheminimumofthe$$$$functionf\left(x\right)=x^3+\frac{5}{x}.$$

Answered by Frix last updated on 16/Jul/23

$$f^{\prime }\left(x\right)=0$$$$3x^2-\frac{5}{x^2}=0$$$$x=\pm {\left(\frac{5}{3}\right)}^{\frac{1}{4}}$$$$f\left({\left(\frac{5}{3}\right)}^{\frac{1}{4}}\right)=4\times {\left(\frac{5}{3}\right)}^{\frac{3}{4}}$$

Answered by mr W last updated on 16/Jul/23

$$anotherway:$$$$f\left(x\right)=x^3+\frac{5}{3x}+\frac{5}{3x}+\frac{5}{3x}$$$$⩾4{\left(x^3\times \frac{5}{3x}\times \frac{5}{3x}\times \frac{5}{3x}\right)}^{\frac{1}{4}}=4{\left(\frac{5}{3}\right)}^{\frac{3}{4}}$$$$\Rightarrow minimum=4{\left(\frac{5}{3}\right)}^{\frac{3}{4}}$$$$when{x}^3=\frac{5}{3x},i.e.x={\left(\frac{5}{3}\right)}^{\frac{1}{4}}$$

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