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Question Number 170576 by Tawa11 last updated on 27/May/22

Commented by mr W last updated on 27/May/22

f′(x)=3x^2 +(1/x^2 )>0 f(x) is strictly increasing. f(−1)=−1−2+1=−2 <0 f(0^− )→+∞ >0 ⇒there must be a zero in (−1,0) f(1)=1−2−1=−2 <0 f(2)=8−2−(1/2)=((11)/2) >0 ⇒there must be a zero in (1,2)

$${f}'\left({x}\right)=\mathrm{3}{x}^{\mathrm{2}} +\frac{\mathrm{1}}{{x}^{\mathrm{2}} }>\mathrm{0} \\ $$$${f}\left({x}\right)\:{is}\:{strictly}\:{increasing}. \\ $$$${f}\left(−\mathrm{1}\right)=−\mathrm{1}−\mathrm{2}+\mathrm{1}=−\mathrm{2}\:<\mathrm{0} \\ $$$${f}\left(\mathrm{0}^{−} \right)\rightarrow+\infty\:>\mathrm{0} \\ $$$$\Rightarrow{there}\:{must}\:{be}\:{a}\:{zero}\:{in}\:\left(−\mathrm{1},\mathrm{0}\right) \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{1}−\mathrm{2}−\mathrm{1}=−\mathrm{2}\:<\mathrm{0} \\ $$$${f}\left(\mathrm{2}\right)=\mathrm{8}−\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{11}}{\mathrm{2}}\:>\mathrm{0} \\ $$$$\Rightarrow{there}\:{must}\:{be}\:{a}\:{zero}\:{in}\:\left(\mathrm{1},\mathrm{2}\right) \\ $$

Commented by Tawa11 last updated on 27/May/22

I really appreciate sir.

$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir}. \\ $$

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