Choosing the Correct Branch for an Inverse
An engineer is calibrating a sensor whose response is modeled by f(x) = x^2 − 4x + 7, but the sensor is only designed to accept inputs x ≥ 2. Two technicians propose inverse functions for converting a reading back to an input. Technician A proposes f⁻¹(x) = 2 + √(x − 3); Technician B proposes f⁻¹(x) = 2 − √(x − 3), arguing that both signs are needed because squaring loses information. A test reading is f(5) = 12. Plugging 12 into the two proposals gives 5 and −1, respectively. The engineer notes that the sensor cannot have accepted x = −1 and asks for a decision based on the restricted domain, not on the algebraic formula alone. The group also wants to know the correct domain and range of the inverse before using it.
1. Rewrite f in vertex form, state its domain and range under the sensor's restriction, and explain why an inverse exists.
Read the worked answer
Complete the square: f(x) = x^2 − 4x + 7 = (x − 2)^2 + 3. The vertex is (2, 3). On the restricted domain x ≥ 2, the graph is the right-hand branch of the parabola and is strictly increasing. Therefore f is one-to-one on that domain and has an inverse. The range is [3, ∞). The inverse's domain must equal the original range, [3, ∞), and the inverse's range must equal the original domain, [2, ∞). Without the restriction x ≥ 2, the parabola would not be one-to-one, because outputs above 3 would come from two different inputs.
2. Find f⁻¹(x), verify it using f(5) = 12, and explain why Technician B's branch must be rejected.
Read the worked answer
Let y = (x − 2)^2 + 3. Then y − 3 = (x − 2)^2. Since x ≥ 2, the quantity x − 2 is nonnegative, so x − 2 = +√(y − 3), not the negative root. Thus x = 2 + √(y − 3). Swapping the variable names gives f⁻¹(x) = 2 + √(x − 3), with domain x ≥ 3. Check: f(5) = 25 − 20 + 7 = 12, and f⁻¹(12) = 2 + √9 = 5. Technician B's value at 12 is 2 − 3 = −1, which is outside the sensor's input domain x ≥ 2 and would have been impossible. The negative branch corresponds to the left-hand branch x ≤ 2, which is not the model's domain.
3. A colleague computes f⁻¹(f(0)) = 4 and says the inverse has recovered the original input. Is this valid? Explain using the restricted domain.
Read the worked answer
No. The sensor model only accepts inputs x ≥ 2, so f(0) is not defined in this model; 0 is outside the restricted domain. If one ignores the restriction and uses the unrestricted parabola, f(0) = 7 and f⁻¹(7) = 2 + √4 = 4, but 4 is not the original input 0. The unrestricted parabola is not one-to-one: f(0) = f(4) = 7. The restricted inverse returns the unique input in [2, ∞) that produces output 7, which is 4. It cannot recover 0 because 0 was never allowed. The computation is invalid unless the model is explicitly extended to include x = 0, which would destroy invertibility.