Найдём нули подмодульных выражений:
![x-3=0; x=3 x-3=0; x=3](https://tex.z-dn.net/?f=+x-3%3D0%3B+x%3D3+)
![3x-2=0; x=\frac{2}{3} 3x-2=0; x=\frac{2}{3}](https://tex.z-dn.net/?f=+3x-2%3D0%3B+x%3D%5Cfrac%7B2%7D%7B3%7D++)
Проставим знаки, подставляя значения из промежутков в модули (см. рис.)
![1) x<\frac{2}{3} 1) x<\frac{2}{3}](https://tex.z-dn.net/?f=+1%29+x%3C%5Cfrac%7B2%7D%7B3%7D++)
![y=2(3-x)+2-3x=6-2x+2-3x=-5x+8 y=2(3-x)+2-3x=6-2x+2-3x=-5x+8](https://tex.z-dn.net/?f=+y%3D2%283-x%29%2B2-3x%3D6-2x%2B2-3x%3D-5x%2B8+)
![y=5x-8, x<\frac{2}{3} y=5x-8, x<\frac{2}{3}](https://tex.z-dn.net/?f=+y%3D5x-8%2C+x%3C%5Cfrac%7B2%7D%7B3%7D++)
![2) \frac{2}{3}\leq x\leq 3 2) \frac{2}{3}\leq x\leq 3](https://tex.z-dn.net/?f=+2%29+%5Cfrac%7B2%7D%7B3%7D%5Cleq+x%5Cleq+3+)
![y=2(3-x)+3x-2=6-2x+3x-2=x+4, \frac{2}{3} \leq x\leq 3 y=2(3-x)+3x-2=6-2x+3x-2=x+4, \frac{2}{3} \leq x\leq 3](https://tex.z-dn.net/?f=+y%3D2%283-x%29%2B3x-2%3D6-2x%2B3x-2%3Dx%2B4%2C+%5Cfrac%7B2%7D%7B3%7D+%5Cleq+x%5Cleq+3+)
![y=x+4, \frac{2}{3} \leq x\leq 3 y=x+4, \frac{2}{3} \leq x\leq 3](https://tex.z-dn.net/?f=+y%3Dx%2B4%2C+%5Cfrac%7B2%7D%7B3%7D+%5Cleq+x%5Cleq+3+)
3 " alt=" 3) x>3 " align="absmiddle" class="latex-formula">
3 " alt=" y=2(x-3)+3x-2=2x-6+3x-2=5x-8, x>3 " align="absmiddle" class="latex-formula">
3 " alt=" y=5x-8, x>3 " align="absmiddle" class="latex-formula">
То есть функция убывает до
, а после возрастает
Подставляем ![x=\frac{2}{3} x=\frac{2}{3}](https://tex.z-dn.net/?f=+x%3D%5Cfrac%7B2%7D%7B3%7D++)
![y(\frac{2}{3})=2|\frac{2}{3}-3|+|\frac{3*2}{3}-2|=2|3-\frac{2}{3}|=\frac{14}{3} y(\frac{2}{3})=2|\frac{2}{3}-3|+|\frac{3*2}{3}-2|=2|3-\frac{2}{3}|=\frac{14}{3}](https://tex.z-dn.net/?f=+y%28%5Cfrac%7B2%7D%7B3%7D%29%3D2%7C%5Cfrac%7B2%7D%7B3%7D-3%7C%2B%7C%5Cfrac%7B3%2A2%7D%7B3%7D-2%7C%3D2%7C3-%5Cfrac%7B2%7D%7B3%7D%7C%3D%5Cfrac%7B14%7D%7B3%7D++)
Ответ: ![\frac{14}{3} \frac{14}{3}](https://tex.z-dn.net/?f=+%5Cfrac%7B14%7D%7B3%7D++)