Question Number 171284 by Kodjo last updated on 11/Jun/22
$${g}\left({x}\right)=−{x}^{\mathrm{2}} +\mathrm{1}−{ln}\mid{x}\mid \\ $$Study the variations of the function g and draw up its table of variations
Answered by a.lgnaoui last updated on 12/Jun/22
Answered by a.lgnaoui last updated on 12/Jun/22
$${pour}\:{x}<\mathrm{0}\:\:\:{g}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{1}−{ln}\left(−{x}\right) \\ $$$${g}'\left({x}\right)=\mathrm{2}{x}+\frac{\mathrm{1}}{{x}}=\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}}{{x}}<\mathrm{0}\:{g}\left({x}\right)\:{decroissant} \\ $$$${x}\:\:\:\:\:−\infty\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:+\infty\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\backslash\:\:\:\:\:\:\:\:\:\:−\infty \\ $$$${limg}\left({x}\right)_{{x}\rightarrow−\infty} =\mathrm{2}{x}−\frac{\mathrm{1}}{−{x}}=\mathrm{2}{x}+\frac{\mathrm{1}}{{x}}=+\infty \\ $$$${limg}\left({x}\right)_{{x}\rightarrow−\mathrm{0}} =−\infty \\ $$$$ \\ $$