If f(x) = frac{x}{x^2+1} is an increasing fun | vedclass

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(D) Given $f(x) = \frac{x}{x^2+1}$.To find the interval where $f(x)$ is increasing,we find its derivative $f'(x)$ using the quotient rule:$f'(x) = \fr

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$(-1, 1)$

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(D) Given $f(x) = \frac{x}{x^2+1}$.To find the interval where $f(x)$ is increasing,we find its derivative $f'(x)$ using the quotient rule:$f'(x) = \frac{(x^2+1)(1) - x(2x)}{(x^2+1)^2} = \frac{x^2+1-2x^2}{(x^2+1)^2} = \frac{1-x^2}{(x^2+1)^2}$.For $f(x)$ to be an increasing function,we must have $f'(x) > 0$.Since $(x^2+1)^2$ is always positive for all real $x$,the condition $f'(x) > 0$ implies $1-x^2 > 0$.This simplifies to $x^2 - 1 Solving this inequality,we get $x \in (-1, 1)$.

If $f(x) = \frac{x}{x^2+1}$ is an increasing function,then the value of $x$ lies in:

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