Use b(x)=x2\displaystyle {b}{\left({x}\right)}={x}^{{2}}b(x)=x2 and j(x)=x−3\displaystyle {j}{\left({x}\right)}={x}-{3}j(x)=x−3 to solve:
(b∘j)(x)=(j∘b)(x)\displaystyle {\left({b}\circ{j}\right)}{\left({x}\right)}={\left({j}\circ{b}\right)}{\left({x}\right)}(b∘j)(x)=(j∘b)(x)
x=\displaystyle {x}=x=
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