Let f : R to R be a function defined by | vedclass

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Question

$f(x)=-\frac{|x|^3+|x|}{1+x^2}$

$f(-x)=-\frac{|-x|^3+|-x|}{1+(-x)^2}=-\frac{|x|^3+|x|}{1+x^2}=-f(x)$

Clearly, $f ( x )$ is an even function and

Vedclass · Accepted Answer

$III$ and $IV$ Quadrants

Vedclass · Answer

$f(x)=-\frac{|x|^3+|x|}{1+x^2}$

$f(-x)=-\frac{|-x|^3+|-x|}{1+(-x)^2}=-\frac{|x|^3+|x|}{1+x^2}=-f(x)$

Clearly, $f ( x )$ is an even function and $f ( x ) < 0$ for all $x > 0$.

Therefore, the graph of $f(x)$ lies in the third and fourth quadrant.

Let $f : R \to R$ be a function defined by $f(x) = - \frac{{|x{|^3} + |x|}}{{1 + {x^2}}}$; then the graph of $f(x)$ is lies in the :-

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