Transformations and the Missing Domain
A student is asked to find the inverse of f(x) = x^2 - 4x + 7 for x >= 2. They complete the square to get f(x) = (x-2)^2 + 3, then write f^{-1}(x) = 2 + sqrt(x - 3), stating the domain as all real x. A classmate says the inverse should be 2 - sqrt(x-3) instead. The teacher asks the pair to justify both the formula and the domain, noting that a function and its inverse must swap domain and range. The graph of f is a parabola with vertex at (2,3).
1. Determine the correct rule and domain for f^{-1}, and adjudicate between the two students.
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For x >= 2, f(x) = (x-2)^2 + 3 has range [3, infinity) since the squared term is non-negative. The inverse must have domain [3, infinity) and range [2, infinity). Solving y = (x-2)^2 + 3 gives x - 2 = +/- sqrt(y - 3); the restriction x >= 2 forces the positive root, so f^{-1}(x) = 2 + sqrt(x - 3). The first student's formula is right but the domain 'all real x' is wrong: sqrt(x-3) requires x >= 3. The classmate's minus sign corresponds to the branch x <= 2, which is not the given domain. So the first student wins on formula but loses on domain; the second is wrong entirely for this restriction.
2. Explain, using the vertex, why choosing x >= 2 rather than x <= 2 matters, and what would change if the domain were x <= 2.
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The vertex at (2,3) is the turning point. Restricting to x >= 2 selects the right-hand branch, on which f is strictly increasing, hence one-to-one and invertible. Restricting to x <= 2 selects the left branch, also one-to-one but decreasing, and the inverse becomes f^{-1}(x) = 2 - sqrt(x - 3) with domain [3, infinity) and range (-infinity, 2]. The key judgement is that a quadratic is not invertible on its full domain; the chosen restriction determines both the sign of the root and the range of the inverse. Students who ignore the restriction produce a relation, not a function.
3. Verify the inverse by composing f and f^{-1}, and state what the composition should equal.
Read the worked answer
For x >= 3: f(f^{-1}(x)) = ((2 + sqrt(x-3)) - 2)^2 + 3 = (sqrt(x-3))^2 + 3 = (x - 3) + 3 = x. For x >= 2: f^{-1}(f(x)) = 2 + sqrt((x-2)^2 + 3 - 3) = 2 + sqrt((x-2)^2) = 2 + |x - 2|. Since x >= 2, |x - 2| = x - 2, giving 2 + (x - 2) = x. Both compositions return the identity on the appropriate domains. The absolute-value step is where careless answers fail: sqrt((x-2)^2) is |x-2|, not x-2, and only the domain restriction licenses dropping the absolute value.