Open-circuit Output Voltage of Bipolar Output Stage
Notice how at \$Io \gtrapprox 0\$, Vo drops sharply from Vcc down to 3.5V. This is the output voltage with just enough load to bias Q1+Q2 so \$ V_o = V_{cc} - R_1 \cdot I(R_1) - V_{be}(Q_1) - V_{be}(Q_2)\$ with \$I(R_1) = I_b(Q_1) \approx 0\$. That's a bit less than 5v - 0.7v - 0.7v = 3.6v.