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

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

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Decreasing for $x  0$

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(C) Given the function $f(x) = 1 - e^{-x^2/2}$.To determine the intervals of increase and decrease,we find the derivative $f'(x)$:$f'(x) = \frac{d}{dx}(1 - e^{-x^2/2}) = 0 - e^{-x^2/2} \cdot \frac{d}{dx}(-x^2/2) = -e^{-x^2/2} \cdot (-x) = x e^{-x^2/2}$.Since $e^{-x^2/2}$ is always positive for all real $x$,the sign of $f'(x)$ depends solely on $x$.$1$. For $f(x)$ to be increasing,we require $f'(x) > 0$,which implies $x e^{-x^2/2} > 0$. Since $e^{-x^2/2} > 0$,this holds when $x > 0$.$2$. For $f(x)$ to be decreasing,we require $f'(x)  0$,this holds when $x Therefore,the function is decreasing for $x  0$.

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

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