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Question Number 171284 by Kodjo last updated on 11/Jun/22

g(x)=−x^2 +1−ln∣x∣

$${g}\left({x}\right)=−{x}^{\mathrm{2}} +\mathrm{1}−{ln}\mid{x}\mid \\ $$ Study the variations of the function g and draw up its table of variations\\n\\n

Answered by a.lgnaoui last updated on 12/Jun/22

Answered by a.lgnaoui last updated on 12/Jun/22

pour x<0 g(x)=x^2 +1−ln(−x) g′(x)=2x+(1/x)=((2x^2 +1)/x)<0 g(x) decroissant x −∞ 0 +∞ \ −∞ limg(x)_(x→−∞) =2x−(1/(−x))=2x+(1/x)=+∞ limg(x)_(x→−0) =−∞

$${pour}\:{x}<\mathrm{0}\:\:\:{g}\left({x}\right)={x}^{\mathrm{2}} +\mathrm{1}−{ln}\left(−{x}\right) \\ $$ $${g}'\left({x}\right)=\mathrm{2}{x}+\frac{\mathrm{1}}{{x}}=\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}}{{x}}<\mathrm{0}\:{g}\left({x}\right)\:{decroissant} \\ $$ $${x}\:\:\:\:\:−\infty\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0} \\ $$ $$\:\:\:\:\:\:\:\:+\infty\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\backslash\:\:\:\:\:\:\:\:\:\:−\infty \\ $$ $${limg}\left({x}\right)_{{x}\rightarrow−\infty} =\mathrm{2}{x}−\frac{\mathrm{1}}{−{x}}=\mathrm{2}{x}+\frac{\mathrm{1}}{{x}}=+\infty \\ $$ $${limg}\left({x}\right)_{{x}\rightarrow−\mathrm{0}} =−\infty \\ $$ $$ \\ $$

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