Calibrating a Quadratic Power Model with a Restricted Inverse
A small wind-turbine test rig produces a useful power output modelled by P(v) = v^2 − 6v + k kilowatts for wind speeds v ≥ 3 m/s, where v is measured in m/s and k is a calibration constant. The technician needs the output to have range [1, ∞). A colleague claims that any k > 9 will work because the quadratic is increasing for v ≥ 3; another argues that the vertex must be placed correctly and that the inverse function must exist on the stated domain. The rig's control software only accepts an inverse function when the original function is one-to-one on its domain. During a test, the turbine is run at v = 5 m/s and the displayed output is used to check the calibration. The team also wants to know where the graph of P meets the graph of its inverse, because the controller switches mode at such points. k is unknown, and the team has only one calibration run to settle the value.
1. Determine the value of k that gives range [1, ∞), and state the domain and range of P^{-1}. Show how the vertex condition fixes k.
Read the worked answer
Complete the square: P(v) = (v − 3)^2 + k − 9. On the domain v ≥ 3, the squared term is at least 0 and equals 0 at v = 3, so the minimum output is k − 9. Hence the range is [k − 9, ∞). Requiring this to equal [1, ∞) gives k − 9 = 1, so k = 10. The colleague's claim is wrong: k > 9 gives range [k − 9, ∞), which starts above 1 unless k = 10. With k = 10, P(v) = (v − 3)^2 + 1 is one-to-one on v ≥ 3 because it is strictly increasing there. Therefore P^{-1} has domain [1, ∞) and range [3, ∞).
2. Find P^{-1}(v) and solve P(v) = P^{-1}(v), explaining why the solution can be found from P(v) = v and why any other algebraic root is rejected.
Read the worked answer
From P = (v − 3)^2 + 1, with v ≥ 3, rearrange to (v − 3)^2 = P − 1 and take the positive root: v = 3 + √(P − 1). Thus P^{-1}(v) = 3 + √(v − 1), for v ≥ 1. Because P is strictly increasing and continuous on [3, ∞), any intersection of y = P(v) and y = P^{-1}(v) must lie on y = v. To see this, suppose P(a) = b and P^{-1}(a) = b, so P(b) = a. If a < b, then increasing P gives P(a) < P(b), i.e. b < a, a contradiction; if a > b, then P(a) > P(b), i.e. b > a, again a contradiction. Hence a = b. So solve P(v) = v: v^2 − 6v + 10 = v, giving v^2 − 7v + 10 = 0, so (v − 2)(v − 5) = 0. The root v = 2 is outside the domain v ≥ 3 and is rejected. The valid intersection is v = 5, and checking gives P(5) = 5 and P^{-1}(5) = 3 + √4 = 5.