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Question Number 121202 by benjo_mathlover last updated on 05/Nov/20

Commented by liberty last updated on 05/Nov/20

$$\mathrm{g}\left(\mathrm{x}\right)=\frac{{\mathrm{x}}^2+\mathrm{x}-1}{\mathrm{x}}=\mathrm{x}+1-\frac{1}{\mathrm{x}}$$$${\mathrm{g}}^{\prime }\left(\mathrm{x}\right)=1+\frac{1}{{\mathrm{x}}^2}>0,\mathrm{g}\left(\mathrm{x}\right)\mathrm{increasing}\mathrm{both}\mathrm{sides}$$$$\mathrm{interval}(-\mathrm{\infty },0)\cup (0,\mathrm{\infty }).\mathrm{Then}\mathrm{g}\left(\mathrm{x}\right)$$$$\mathrm{no}\mathrm{have}\mathrm{local}\mathrm{maxima}\mathrm{and}\mathrm{local}\mathrm{minima}$$

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