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

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(D) Given the function $f(x) = - \frac{|x|^3 + |x|}{1 + x^2}$.First,check for symmetry by evaluating $f(-x)$:$f(-x) = - \frac{|-x|^3 + |-x|}{1 + (-x)^

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$III$ and $IV$ Quadrants

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(D) Given the function $f(x) = - \frac{|x|^3 + |x|}{1 + x^2}$.First,check for symmetry by evaluating $f(-x)$:$f(-x) = - \frac{|-x|^3 + |-x|}{1 + (-x)^2} = - \frac{|x|^3 + |x|}{1 + x^2} = f(x)$.Since $f(-x) = f(x)$,the function is an even function,meaning the graph is symmetric about the $y$-axis.Next,analyze the sign of $f(x)$:For any $x \in R$,$|x| \ge 0$ and $x^2 \ge 0$,so $|x|^3 + |x| \ge 0$ and $1 + x^2 > 0$.Since there is a negative sign in front of the fraction,$f(x) \le 0$ for all $x \in R$.Because $f(x) \le 0$ for all $x$,the graph of the function never enters the first or second quadrants (where $y > 0$).Since the function is defined for all $x \in R$ and $f(x) \le 0$,the graph lies entirely in the third and fourth quadrants.

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

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