fungsi f(x) = x-1,g(x) = 3-2x, dan h(x) = x² +x+1 tentukan = a) (f°h°g) (x) b) (f°h°g) (-3) c) (g°h°f) (x) d) (g°h°f) (5)

a.
[tex](f\circ h \circ g)(x) = f(h(g(x))) \\ f(h(3-2x))=f((3-2x)^2+(3-2x)+1) \\ f(9-12x+4x^2+3-2x+1)=f(4x^2-14x+13) \\ (4x^2-14x+13)-1=4x^2-14x+12[/tex]
b.
Langsung substitusi aja dari yang tadi,
[tex](f\circ h \circ g)(-3)=4(-3)^2-14(-3)+12=36+42+12=90[/tex]
c.
[tex](g\circ h \circ f)(x)=g(h(f(x))) \\ g(h(x-1))=g((x-1)^2+(x-1)+1) \\ g(x^2-2x+1+x-1+1) \\ g(x^2-3x+1)=3-2(x^2-3x+1) \\ 3-2x^2+6x-2 = -2x^2+6x+1[/tex]
d.
Langsung substitusi aja dari yang tadi,
[tex](g\circ h \circ f)(5)=-2.5^2+6.5+1=-50+30+1=-19[/tex]