(b) (f∘g)(x)=f(g(x))=(g(x))2+1=(x+2)2+1=(f \circ g)(x) = f(g(x)) = (g(x))^2+ 1 = (x+2)^2+1 =(f∘g)(x)=f(g(x))=(g(x))2+1=(x+2)2+1= x2+4x+4+1=x2+4x+5x^2 +4x+4+1= x^2+4x+5x2+4x+4+1=x2+4x+5
(g∘f)(x)=g(f(x))=f(x)+2=x2+1+2=x2+3(g \circ f)(x) = g(f(x)) = f(x) +2 = x^2+1+2 = x^2+3(g∘f)(x)=g(f(x))=f(x)+2=x2+1+2=x2+3
(c) (f+g)(x)=f(x)+g(x)=x2+1+x+2=x2+x+3(f+g)(x) = f(x) + g(x) = x^2+1 +x+2 = x^2+x+3(f+g)(x)=f(x)+g(x)=x2+1+x+2=x2+x+3
(fg)(x)=f(x)⋅g(x)=(x2+1)(x+2)=x3+2x2+x+2(fg)(x) = f(x) \cdot g(x) = (x^2+1)(x+2) = x^3 + 2x^2+x+2(fg)(x)=f(x)⋅g(x)=(x2+1)(x+2)=x3+2x2+x+2
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