g(x)=−x^2 +1−ln∣x∣


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

$g (x) = - x^{2} + 1 - l n ∣ x ∣$ 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
	$pour x<0 g(x)=x^2 +1−ln(−x) g′(x)=2x+(1/x)=((2x^2 +1)/x)<0 g(x) decroissant x −∞ 0 +∞ \ −∞ limg(x)_(x→−∞) =2x−(1/(−x))=2x+(1/x)=+∞ limg(x)_(x→−0) =−∞$
	$p o u r x < 0 g (x) = x^{2} + 1 - l n (- x)$ $g^{'} (x) = 2 x + \frac{1}{x} = \frac{2 x^{2} + 1}{x} < 0 g (x) d e c r o i s s a n t$ $x - \infty 0$ $+ \infty ∖ - \infty$ $l i m g {(x)}_{x \to - \infty} = 2 x - \frac{1}{- x} = 2 x + \frac{1}{x} = + \infty$ $l i m g {(x)}_{x \to - 0} = - \infty$
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