If f(x) = intlimits_0^x {{e^{frac{{ - {t^2}}} | vedclass

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(B) Given $f(x) = \int\limits_0^x {{e^{\frac{{ - {t^2}}}{2}}}} \left( {1 - {t^2}} \right)\,dt$. Applying the Leibniz rule for differentiation,we get:

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(B) Given $f(x) = \int\limits_0^x {{e^{\frac{{ - {t^2}}}{2}}}} \left( {1 - {t^2}} \right)\,dt$. Applying the Leibniz rule for differentiation,we get: $f'(x) = e^{\frac{{ - {x^2}}}{2}} (1 - x^2)$. For local extrema,set $f'(x) = 0$: $e^{\frac{{ - {x^2}}}{2}} (1 - x^2) = 0$. Since $e^{\frac{{ - {x^2}}}{2}} > 0$ for all $x$,we have $1 - x^2 = 0$,which implies $x^2 = 1$,so $x = \pm 1$. Now,find the second derivative $f''(x)$: $f''(x) = e^{\frac{{ - {x^2}}}{2}} (-2x) + (1 - x^2) e^{\frac{{ - {x^2}}}{2}} (-x) = e^{\frac{{ - {x^2}}}{2}} (-2x - x + x^3) = e^{\frac{{ - {x^2}}}{2}} (x^3 - 3x)$. Check the sign of $f''(x)$ at critical points: At $x = 1$: $f''(1) = e^{-1/2} (1 - 3) = -2e^{-1/2} At $x = -1$: $f''(-1) = e^{-1/2} (-1 + 3) = 2e^{-1/2} > 0$ (Local Minimum). Therefore,$f(x)$ is minimum at $x = -1$.

If $f(x) = \int\limits_0^x {{e^{\frac{{ - {t^2}}}{2}}}} \left( {1 - {t^2}} \right)\,dt$,then $f(x)$ is minimum at $x = \dots$

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