show that f°f=2f
f=x+1
f(x)=x+1f(x)=x+1f(x)=x+1. The composition f∘ff\circ ff∘f has the form: f∘f(x)=f(x+1)=x+2f\circ f(x)=f(x+1)=x+2f∘f(x)=f(x+1)=x+2;
2f(x)=2x+22f(x)=2x+22f(x)=2x+2. As we can see, f∘f(x)≠2f(x)f\circ f(x)\neq2f(x)f∘f(x)=2f(x).
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