(4) if a function f(x) in the domain x ∈ 0,2 is f(x) = |x - 1|+|x² - 2x|, then the minimum value is 1 - 8…

Explanation:

Step1: Analyze absolute - value expressions

First, rewrite the absolute - value functions. For (y = |x - 1|), when (x\geq1), (y=x - 1); when (x\lt1), (y = 1 - x). For (y=|x^{2}-2x|=|x(x - 2)|), when (x\in[0,2]), (y=-x^{2}+2x).

Step2: Split the domain

Split the domain ([0,2]) into sub - intervals. When (x\in[0,1]), (f(x)=(1 - x)+(-x^{2}+2x)=-x^{2}+x + 1). When (x\in[1,2]), (f(x)=(x - 1)+(-x^{2}+2x)=-x^{2}+3x - 1).

Step3: Find the minimum and maximum on ([0,1])

For (y=-x^{2}+x + 1) on (x\in[0,1]), its derivative (y^\prime=-2x + 1). Let (y^\prime = 0), then (x=\frac{1}{2}). (y(\frac{1}{2})=-\left(\frac{1}{2}\right)^{2}+\frac{1}{2}+1=\frac{-1 + 2+4}{4}=\frac{5}{4}), (y(0)=1), (y(1)=1).

Step4: Find the minimum and maximum on ([1,2])

For (y=-x^{2}+3x - 1) on (x\in[1,2]), its derivative (y^\prime=-2x + 3). Let (y^\prime = 0), then (x=\frac{3}{2}). (y(\frac{3}{2})=-\left(\frac{3}{2}\right)^{2}+3\times\frac{3}{2}-1=\frac{-9 + 18 - 4}{4}=\frac{5}{4}), (y(1)=1), (y(2)=-4 + 6 - 1=1).

Answer:

The minimum value is (1), the maximum value is (\frac{5}{4})