f(x) = intlimits_0^x t(t - 1)(t - 2) dt takes | vedclass

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(C) Given $f(x) = \int\limits_0^x t(t - 1)(t - 2) dt$. By the Fundamental Theorem of Calculus,$f'(x) = x(x - 1)(x - 2)$. To find the critical points,s

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$x = 0, 2$

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(C) Given $f(x) = \int\limits_0^x t(t - 1)(t - 2) dt$. By the Fundamental Theorem of Calculus,$f'(x) = x(x - 1)(x - 2)$. To find the critical points,set $f'(x) = 0$,which gives $x = 0, 1, 2$. We analyze the sign of $f'(x)$ around these points: For $x For $0  0$. For $1 For $x > 2$,$f'(x) > 0$. At $x = 0$,$f'(x)$ changes from negative to positive,so $x = 0$ is a local minimum. At $x = 1$,$f'(x)$ changes from positive to negative,so $x = 1$ is a local maximum. At $x = 2$,$f'(x)$ changes from negative to positive,so $x = 2$ is a local minimum. Thus,the function takes its minimum value at $x = 0$ and $x = 2$.

$f(x) = \int\limits_0^x t(t - 1)(t - 2) dt$ takes on its minimum value when:

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