Use the rubric on the exam page. Award method credit for a correct model carried through with a minor arithmetic error, but check domain exclusions, sign conventions and units explicitly.
The root requires −3≤x≤3, and the denominator excludes −1. The domain is [−3,−1) ∪ (−1,3] . The endpoints ±3 give defined zero outputs.
Slope (4−1)/(80−20)=0.05 V/°C. Thus v(T)=0.05T for 20≤T≤80, output 2.5 V at 50 °C and range [1,4] V . Using 100 °C requires extrapolation beyond the calibration interval.
The parabola includes (−2,4), (0,0) and (3,9), with range [0,9] . The secant slope is (1−4)/(1−(−2))=−1. The quotient equals x+1 only for x≠1, giving a hole at (1,2).
Removing negative charge gives Q=+1.602176634 × 10⁻⁷ C≈+0.1602 μC.
The currents are 1 A and 2 A. Overall Iavg=8/5=1.6 A. The interval durations differ, so the unweighted mean 1.5 A is not the overall average.
VAB=7−(−1)=8 V. The potential-energy decrease is 0.25×8=2 J. An open gap interrupts sustained conduction in the ideal steady-DC model; it does not force source voltage to zero.