The function f(x) = int_1^x { 2(t - 1)(t - 2) | vedclass

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(A) Let $g(t) = 2(t - 1)(t - 2)^3 + 3(t - 1)^2(t - 2)^2$. By the Fundamental Theorem of Calculus,$f'(x) = g(x) = 2(x - 1)(x - 2)^3 + 3(x - 1)^2(x - 2)

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(A) Let $g(t) = 2(t - 1)(t - 2)^3 + 3(t - 1)^2(t - 2)^2$. By the Fundamental Theorem of Calculus,$f'(x) = g(x) = 2(x - 1)(x - 2)^3 + 3(x - 1)^2(x - 2)^2$. Factor out common terms: $f'(x) = (x - 1)(x - 2)^2 [2(x - 2) + 3(x - 1)]$. Simplify the expression: $f'(x) = (x - 1)(x - 2)^2 [2x - 4 + 3x - 3] = (x - 1)(x - 2)^2 (5x - 7)$. To find critical points,set $f'(x) = 0$,which gives $x = 1, x = 2, x = 7/5$. Analyze the sign of $f'(x)$ around these points: For $x  0$. For $1 For $7/5  0$. For $x > 2$,$f'(x) > 0$. At $x = 1$,the derivative changes from positive to negative,so $f(x)$ has a local maximum at $x = 1$.

The function $f(x) = \int_1^x \{ 2(t - 1)(t - 2)^3 + 3(t - 1)^2(t - 2)^2 \} dt$ is maximum when $x$ is equal to

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