Question
show that $(f \circ g)(x)=(g \circ f)(x)=x$$$f(x)=a x+b ; g(x)=\frac{1}{a}(x-b) \quad a \neq 0$$
Step 1
This means we substitute $g(x)$ into $f(x)$. So, we have: $$ (f \circ g)(x)=f(g(x))=f\left(\frac{1}{a}(x-b)\right) $$ Show more…
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