The function 11+x2 is decreasing in the interval

y=11+x2 ⇒ dydx=−2x(1+x2)2To be decreasing, −2x(1+x2) 0 ⇒ x∈(0,∞)

The function 11+x2 is decreasing in the interval

y=11+x2 ⇒ dydx=−2x(1+x2)2To be decreasing, −2x(1+x2) 0 ⇒ x∈(0,∞)

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

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$(- \infty, - 1]$

$(- \infty, 0]$

$[1, \infty)$

$(0, \infty)$

answer is D.

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Detailed Solution

$y = \frac{1}{1 + x^{2}} \Rightarrow \frac{d y}{d x} = - \frac{2 x}{{(1 + x^{2})}^{2}}$

To be decreasing, $- \frac{2 x}{(1 + x^{2})} < 0 \Rightarrow x > 0 \Rightarrow x \in (0, \infty)$

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