The function f(x) = int_{-1}^x {t({e^t} - 1)( | vedclass

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(B) Given $f(x) = \int_{-1}^x {t({e^t} - 1)(t - 1){(t - 2)}^3{(t - 3)}^5} dt$.By the Fundamental Theorem of Calculus,$f'(x) = x({e^x} - 1)(x - 1){(x -

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$1$

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(B) Given $f(x) = \int_{-1}^x {t({e^t} - 1)(t - 1){(t - 2)}^3{(t - 3)}^5} dt$.By the Fundamental Theorem of Calculus,$f'(x) = x({e^x} - 1)(x - 1){(x - 2)}^3{(x - 3)}^5$.For local extrema,we set $f'(x) = 0$,which gives critical points $x = 0, 1, 2, 3$.We examine the sign change of $f'(x)$ around these points:$1$. At $x = 0$: $f'(x)$ changes from negative to negative (no extremum).$2$. At $x = 1$: $f'(x)$ changes from negative to positive (local minimum).$3$. At $x = 2$: $f'(x)$ changes from positive to negative (local maximum).$4$. At $x = 3$: $f'(x)$ changes from negative to positive (local minimum).Thus,the function has local minima at $x = 1$ and $x = 3$. Since $1$ is an option,the correct answer is $1$.

The function $f(x) = \int_{-1}^x {t({e^t} - 1)(t - 1){(t - 2)}^3{(t - 3)}^5} dt$ has a local minimum at $x = $ ..........

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