Let \( P(s) \) be the function \( \frac{s}{s+1}-1 \). Find the following: \[ \begin{array}{l}P(11)= \\ P\left(\frac{1}{8}\right)=\square \\ P\left(-\frac{1}{3}\right)=\square \\ \text { In each box, enter your answer as an integer or reduced fraction. Enter DNE for Does Not Exist, or oo for } \\ \text { Infinity. }\end{array} \]

To find the values of the function \( P(s) = \frac{s}{s+1} - 1 \), we will evaluate it for different values of \( s \). 1. **Calculate \( P(11) \)**: \[ P(11) = \frac{11}{11 + 1} - 1 = \frac{11}{12} - 1 = \frac{11}{12} - \frac{12}{12} = \frac{11 - 12}{12} = \frac{-1}{12} \] 2. **Calculate \( P\left(\frac{1}{8}\right) \)**: \[ P\left(\frac{1}{8}\right) = \frac{\frac{1}{8}}{\frac{1}{8} + 1} - 1 = \frac{\frac{1}{8}}{\frac{1}{8} + \frac{8}{8}} - 1 = \frac{\frac{1}{8}}{\frac{9}{8}} - 1 = \frac{1}{9} - 1 = \frac{1}{9} - \frac{9}{9} = \frac{1 - 9}{9} = \frac{-8}{9} \] 3. **Calculate \( P\left(-\frac{1}{3}\right) \)**: \[ P\left(-\frac{1}{3}\right) = \frac{-\frac{1}{3}}{-\frac{1}{3} + 1} - 1 = \frac{-\frac{1}{3}}{-\frac{1}{3} + \frac{3}{3}} - 1 = \frac{-\frac{1}{3}}{\frac{2}{3}} - 1 = -\frac{1}{2} - 1 = -\frac{1}{2} - \frac{2}{2} = -\frac{3}{2} \] Now we can fill in the boxes: \[ \begin{array}{l} P(11) = \frac{-1}{12} \\ P\left(\frac{1}{8}\right) = \frac{-8}{9} \\ P\left(-\frac{1}{3}\right) = -\frac{3}{2} \\ \end{array} \] So our final answers are: 1. \( P(11) = \frac{-1}{12} \) 2. \( P\left(\frac{1}{8}\right) = \frac{-8}{9} \) 3. \( P\left(-\frac{1}{3}\right) = -\frac{3}{2} \)