Let \(a\) and \(b\) be positive real numbers such that \(a+b=1.\) Find the set of all possible values of \(\frac{1}{a} + \frac{1}{b}.\)
\(\dfrac1{a} + \dfrac1b = \dfrac{a + b}{ab} = \dfrac1{ab}\)
Since a and b are positive real numbers, \(ab > 0\). Consequently, \(\dfrac1{ab} > 0\).
Then, \(\dfrac1a + \dfrac1b \in (0, \infty)\).
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