f(x)

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Първо ще опростим функцията
[tex]f(x)=\frac{x}{5}-\frac{1}{4}(2+3x)+\frac{3-5x}{-20}\Leftrightarrow f(x)=\frac{x}{5}-\frac{2+3x}{4}+\frac{5x-2}{20}\Leftrightarrow f(x)=\frac{4x-5(2+3x)+5x-2}{20}\Leftrightarrow f(x)=\frac{4x-10-15x+5x-2}{20}[/tex]
[tex]\Leftrightarrow f(x)=\frac{-6x-12}{20}\Leftrightarrow f(x)=-\frac{3x+6}{10}[/tex]
a) [tex]f(-2\frac{1}{2})=f(-\frac{5}{2})=-\frac{3.(-\frac{5}{2})+6}{10}=-\frac{-\frac{15}{2}+6}{10}=-\frac{-15+12}{20}=-\frac{-3}{20}=\frac{3}{20}[/tex]
b) [tex]f(1-x)=-\frac{3(1-x)+6}{10}=-\frac{3-3x+6}{10}=-\frac{-3x+9}{10}[/tex]
[tex]f(1-x)=\frac{2-x}{20}\Leftrightarrow -\frac{-3x+9}{10}=\frac{2-x}{20}\Leftrightarrow \frac{3x-9}{10}=\frac{2-x}{20}\Leftrightarrow \frac{6x-18}{20}=\frac{2-x}{20}\Leftrightarrow 6x-18=2-x\Leftrightarrow 7x=20\Leftrightarrow x=\frac{20}{7}[/tex]
c) [tex]f(\frac{x}{6})=-\frac{3.\frac{x}{6}+6}{10}=-\frac{\frac{x}{2}+6}{10}=-\frac{x+12}{20}[/tex]
Неравенството става
[tex]-3(-\frac{x+12}{20})-\frac{3x-1}{2}\ge 0\Leftrightarrow \frac{3x+36}{20}-\frac{30x-10}{20}\ge 0\Leftrightarrow \frac{-27x+46}{20}\ge 0\Leftrightarrow 27x\le 46\Leftrightarrow x\le \frac{46}{27}[/tex]. Тъй като [tex]1<\frac{46}{27}<2[/tex] има само едно естествено число, което удовлетворява неравенството и то е [tex]x=1[/tex].