Należy policzyć wartości funkcji dla argumentów z dziedziny, a z otrzymanych wyników powstanie zbiór wartości.
a)
[tex]f(-4)=f(-3)=f(-2)=f(-1)=f(0)=f(1)=\sqrt3\\ZW_f=\{\sqrt3\}[/tex]
b)
[tex]f(-8)=3-2*|-8|=3-2*8=3-16=-13\\f(-2)=3-2*|-2|=3-2*2=3-4=-1\\f(3)=3-2*|3|=3-2*3=3-6=-3\\ZW-f=\{-3,-1,13\}[/tex]
c)
[tex]f(1)=\sqrt{2*1-1}=\sqrt{1}=1\\f(5)=\sqrt{2*5-1}=\sqrt{9}=3\\f(13)=\sqrt{2*13-1}=\sqrt{25}=5\\f(25)=\sqrt{2*25-1}=\sqrt{49}=7\\ZW_f=\{1,3,5,7\}[/tex]
d)
[tex]f(-4)=\frac{-4-4}{2*(-4)}=\frac{-8}{-8}=1\\f(-2)=\frac{-2-4}{2*(-2)}=\frac{-6}{-4}=\frac{3}{2}=1\frac{1}{2}\\f(1)=\frac{1-4}{2*1}=\frac{-3}{2}=-\frac{3}{2}=-1\frac{1}{2}\\f(2)=\frac{2-4}{2*2}=\frac{-2}{4}=-\frac{1}{2}\\f(3)=\frac{3-4}{2*3}=\frac{-1}{6}=-\frac{1}{6}\\ZW_f=\{-1\frac{1}{2},-\frac{1}{2},-\frac{1}{6},1,1\frac{1}{2}\}[/tex]
e)
[tex]f(\frac{1}{4})=\frac{3-\log_4\frac{1}{4}}{5}=\frac{3-(-1)}{5}=\frac{4}{5}\\f(\frac{1}{2})=\frac{3-\log_4\frac{1}{2}}{5}=\frac{3-(-\frac{1}{2})}{5}=\frac{3\frac{1}{2}}{5}=\frac{7}{2}*\frac{1}{5}=\frac{7}{10}\\f(1)=\frac{3-\log_41}{5}=\frac{3-0}{5}=\frac{3}{5}\\f(2)=\frac{3-\log_42}{5}=\frac{3-\frac{1}{2}}{5}=\frac{2\frac{1}{2}}{5}=\frac{5}{2}*\frac{1}{5}=\frac{1}{2}\\f(4)=\frac{3-\log_44}{5}=\frac{3-1}{5}=\frac{2}{5}\\ZW_f=\{\frac{2}{5},\frac{1}{2},\frac{3}{5},\frac{7}{10},\frac{4}{5}\}[/tex]
f)
[tex]f(-1)=4-6*3^{-1-2}=4-6*3^{-3}=4-6*\frac{1}{27}=4-\frac{2}{9}=3\frac{7}{9}\\f(0)=4-6*3^{0-2}=4-6*3^{-2}=4-6*\frac{1}{9}=4-\frac{2}{3}=3\frac{1}{3}\\f(1)=4-6*3^{1-2}=4-6*3^{-1}=4-6*\frac{1}{3}=4-2=2\\f(2)=4-6*3^{2-2}=4-6*3^{0}=4-6*1=4-6=-2\\f(3)=4-6*3^{3-2}=4-6*3^{1}=4-6*3=4-18=-14\\ZW_f=\{-14,-2,2,3\frac{1}{3},3\frac{7}{9}\}[/tex]
