The function f(x) = (x - 2)|x - 3| is monoton | vedclass

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Question

(A) The function is defined as $f(x) = \begin{cases} (x-2)(x-3) = x^2 - 5x + 6, & x \ge 3 \ (x-2)(-(x-3)) = -(x^2 - 5x + 6) = -x^2 + 5x - 6, & x < 3

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$(-\infty, \frac{5}{2}) \cup (3, \infty)$

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(A) The function is defined as $f(x) = \begin{cases} (x-2)(x-3) = x^2 - 5x + 6, & x \ge 3 \ (x-2)(-(x-3)) = -(x^2 - 5x + 6) = -x^2 + 5x - 6, & x For $x \ge 3$,$f'(x) = 2x - 5$. Since $x \ge 3$,$2x - 5 \ge 2(3) - 5 = 1 > 0$. Thus,$f(x)$ is increasing on $[3, \infty)$.For $x  0$,which means $-2x + 5 > 0$,or $x Combining these intervals,the function is monotonically increasing in $(-\infty, \frac{5}{2}) \cup (3, \infty)$.

The function $f(x) = (x - 2)|x - 3|$ is monotonically increasing in:

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