You can put this solution on YOUR website! If f(x)=x^2-x and g(x)=-3+x find the following:
(a) f(g(x)) (b) g(f(x)) (c) f(f(-1)) (d) g(g(-3))
f(x)=x^2-x ----(1)
g(x)=-3+x ----(2)
(a) f(g(x)) = f(-3+x) using (2)
= f(x-3) (using additive commutativity we have (-3+x)= (x-3) )
=f(t) where t = (x-3)----(*)
= t^2-t using (1)
= (x-3)^2-(x-3)using(*)
=x^2-6x+9-x+3
=x^2-6x-x+9+3
=x^2-7x+12
b)g(f(x)) = g(x^2-x) using (1)
=g(u) where u = x^2-x ----(**)
= -3+u using (2)
= -3+ (x^2-x) using(**)
= x^2-x-3
c) f(f(-1))
= f[(-1)^2-(-1)] (using x = -1 in (1)which is f(x)=(x^2-x))
=f[1+1]
=f(2)
= [(2)^2-(2)] (using x = 2 in (1)which is f(x)=(x^2-x))
= 4-2
=2
(d) g(g(-3))
= g[-3+(-3)] using x = -3 in (2) which is g(x) = -3+x )
=g(-3-3)
=g(-6)
=[-3+(-6)] using x = -6 in (2) which is g(x) = -3+x )
= (-3-6)
= -9
Answer: a)f(g(x)) = x^2-7x+12
b)g(f(x)) = x^2-x-3
c) f(f(-1))=2
(d) g(g(-3))= -9
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