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Answer:
let inverse f(x) be m:
[tex]m = \frac{1}{2x + 1} \\ 2x + 1 = \frac{1}{m} \\ 2x = \frac{1 - m}{m} \\ x = \frac{1 - m}{2m} [/tex]
substitute x in place of m:
[tex]{ \bf{ {f}^{ - 1}(x) = \frac{1 - x}{2x } }}[/tex]