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

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(C) Given the function $f(x) = 1 - e^{-\frac{x^2}{2}}$.To determine the intervals of increase and decrease,we find the derivative $f'(x)$:$f'(x) = \fr

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decreasing for $x  0$.

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(C) Given the function $f(x) = 1 - e^{-\frac{x^2}{2}}$.To determine the intervals of increase and decrease,we find the derivative $f'(x)$:$f'(x) = \frac{d}{dx}(1 - e^{-\frac{x^2}{2}}) = 0 - e^{-\frac{x^2}{2}} \cdot \frac{d}{dx}(-\frac{x^2}{2}) = -e^{-\frac{x^2}{2}} \cdot (-x) = x e^{-\frac{x^2}{2}}$.Since $e^{-\frac{x^2}{2}} > 0$ for all $x \in R$,the sign of $f'(x)$ depends only on the sign of $x$.Case $1$: If $x > 0$,then $f'(x) > 0$,which means the function $f(x)$ is increasing.Case $2$: If $x Therefore,the function is decreasing for $x  0$.

The function $f(x) = 1 - e^{-\frac{x^2}{2}}$ is .......

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