[tex]\boldsymbol{f(x) = 3x-2, \ \ g(x)=2x+3, \ \ h(x) = \dfrac{x+2}{3}} \\[/tex]
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[tex](f \circ g)(x) = f(g(x)) = 3 \cdot g(x) - 2 = 3(2x+3) - 2 = 6x + 9 - 2 = \bf 6x + 7\\[/tex]
[tex](f \circ h)(x) = f(h(x)) = 3 \cdot h(x) - 2 = 3 \cdot \dfrac{x+2}{3} - 2 = x + 2 - 2 = \bf x\\[/tex]
[tex](h \circ f)(x) = h(f(x)) = \dfrac{f(x)+2}{3} = \dfrac{3x - 2 + 2}{3} = \dfrac{3x}{3} = \bf x\\[/tex]
[tex](g \circ h)(x) = g(h(x)) = 2 \cdot h(x) + 3 = 2 \cdot \dfrac{x+2}{3} + 3 = \dfrac{2x+4+9}{3} = \bf \dfrac{2x+13}{3}\\[/tex]
[tex](g \circ f)(x) = g(f(x)) = 2 \cdot f(x) + 3 = 2 \cdot (3x - 2) + 3 = 6x - 4 + 3 = \bf 6x - 1\\[/tex]
[tex](h \circ g)(x) = h(g(x)) = \dfrac{g(x)+2}{3} = \dfrac{2x + 3 + 2}{3} = \bf \dfrac{2x + 5}{3}\\[/tex]
