The function f(x) = frac{1}{1 + x^2} is decre | vedclass

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(D) Let $f(x) = \frac{1}{1 + x^2}$.To find the interval where the function is decreasing,we find its derivative $f'(x)$.Using the chain rule,$f'(x) =

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$(0, \infty )$

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(D) Let $f(x) = \frac{1}{1 + x^2}$.To find the interval where the function is decreasing,we find its derivative $f'(x)$.Using the chain rule,$f'(x) = -\frac{1}{(1 + x^2)^2} \cdot (2x) = -\frac{2x}{(1 + x^2)^2}$.$A$ function is decreasing when $f'(x) So,$-\frac{2x}{(1 + x^2)^2} Since $(1 + x^2)^2$ is always positive for all real $x$,the sign of $f'(x)$ depends only on $-2x$.Therefore,$-2x  0$.Thus,the function is decreasing in the interval $(0, \infty )$.

The function $f(x) = \frac{1}{1 + x^2}$ is decreasing in the interval

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