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Question-121202




Question Number 121202 by benjo_mathlover last updated on 05/Nov/20
Commented by liberty last updated on 05/Nov/20
g(x)=((x^2 +x−1)/x) = x+1−(1/x)  g′(x)=1+(1/x^2 ) > 0 , g(x) increasing both sides  interval (−∞,0) ∪(0,∞) . Then g(x)  no have local maxima and local minima
$$\mathrm{g}\left(\mathrm{x}\right)=\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{x}−\mathrm{1}}{\mathrm{x}}\:=\:\mathrm{x}+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}} \\ $$$$\mathrm{g}'\left(\mathrm{x}\right)=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\:>\:\mathrm{0}\:,\:\mathrm{g}\left(\mathrm{x}\right)\:\mathrm{increasing}\:\mathrm{both}\:\mathrm{sides} \\ $$$$\mathrm{interval}\:\left(−\infty,\mathrm{0}\right)\:\cup\left(\mathrm{0},\infty\right)\:.\:\mathrm{Then}\:\mathrm{g}\left(\mathrm{x}\right) \\ $$$$\mathrm{no}\:\mathrm{have}\:\mathrm{local}\:\mathrm{maxima}\:\mathrm{and}\:\mathrm{local}\:\mathrm{minima} \\ $$

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