Calibrating a Pressure Sensor with a Composite Function
A pressure sensor produces a raw voltage signal modeled by f(t)=(2t+6)/(t-1), where t is elapsed time in seconds and t>=0. The sensor is connected to a display that converts voltage x to pressure reading P(x)=sqrt(x-4) in kilopascals, but the display only accepts voltages for which the square root is defined. A technician tests the system at t=0.5, t=1.5, and t=6 seconds and records whether the display gives a valid reading. The manufacturer also wants an inverse function that recovers elapsed time from a valid voltage, so that the logger can timestamp readings. The technician worries that the composition P(f(t)) may have a smaller domain than either function alone, and that the inverse may not be valid over the entire voltage range.
1. Find the domain of the composite function P(f(t)), simplify the composite expression, and decide which of the three test times produce a valid reading.
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P(x) requires x >= 4, so the composite requires f(t) >= 4. Also t >= 0 and t != 1.
For 0 <= t < 1, the numerator 2t + 6 is positive and the denominator t - 1 is negative, so f(t) < 0, which cannot be at least 4. For t > 1, solve
(2t + 6)/(t - 1) >= 4.
Because t - 1 > 0, multiply without reversing: 2t + 6 >= 4t - 4, so 10 >= 2t, or t <= 5. Thus the domain is 1 < t <= 5.
The simplified composite is
P(f(t)) = sqrt((2t + 6)/(t - 1) - 4) = sqrt((10 - 2t)/(t - 1)) = sqrt(2(5 - t)/(t - 1)).
At t = 0.5, f is negative, so the reading is invalid. At t = 1.5, f(1.5) = 18, which is at least 4, so the reading is valid. At t = 6, f(6) = 3.6, which is less than 4, so the reading is invalid. Only t = 1.5 is valid among the three tests.
2. Find f^{-1}(y) and state the domain on which it is actually usable for this pressure display.
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Start with y = (2t + 6)/(t - 1). Multiplying by t - 1 gives y(t - 1) = 2t + 6, so yt - y = 2t + 6. Rearranging, t(y - 2) = y + 6, hence
f^{-1}(y) = (y + 6)/(y - 2), with y != 2.
For the pressure display, the input voltage must satisfy y >= 4. On that domain the inverse gives t = (y + 6)/(y - 2). As y ranges from 4 upward, t ranges from 5 downward toward 1, which matches the composite domain 1 < t <= 5. So the usable inverse domain is y >= 4, not all real numbers except 2.
3. A second technician claims the inverse can timestamp any voltage above 2. Evaluate that claim.
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The algebraic inverse exists for all y != 2, but the display does not produce a valid pressure unless y >= 4. For a voltage such as y = 3, the inverse formula gives t = (3 + 6)/(3 - 2) = 9 seconds. However, f(9) = 24/8 = 3, and P(3) is not defined because 3 - 4 is negative. Thus y = 3 corresponds to a time when the composition is invalid. Any y in (2, 4) maps to t > 5, outside the composite domain. Therefore the claim is false: the inverse can timestamp only voltages y >= 4 for this display, not every voltage above 2.