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show that f o f^-1(x) = x for y = mx+c
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\[f\circ f^{-1}(x)=x\] by definition of \[f^{-1}\]
but i guess in your case you are supposed to find \[f^{-1}(x)\] and then check
\[y = mx+b\] solve for x get \[x=\frac{y-b}{m}\] and so \[f^{-1}(x)=\frac{x-b}{m}\] and now we can check directly that \[f\circ f^{-1}(x)=x\]
\[f\ circ f^{-1}(x)=f(f^{-1}(x))=f(\frac{x-b}{m})=m\frac{x-b}{m} +b\] and you are done in one algebra step
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