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

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**Step 1:** Evaluate $$ P(11) $$.

$$
P(11)=\frac{11}{11+1} - 1 = \frac{11}{12} - 1 = \frac{11-12}{12} = -\frac{1}{12}
$$

**Step 2:** Evaluate $$ P\left(\frac{1}{8}\right) $$.

$$
P\left(\frac{1}{8}\right)=\frac{\frac{1}{8}}{\frac{1}{8}+1} - 1 = \frac{\frac{1}{8}}{\frac{9}{8}} - 1 = \frac{1}{9} - 1 = \frac{1-9}{9} = -\frac{8}{9}
$$

**Step 3:** Evaluate $$ P\left(-\frac{1}{3}\right) $$.

$$
P\left(-\frac{1}{3}\right)=\frac{-\frac{1}{3}}{-\frac{1}{3}+1} - 1 = \frac{-\frac{1}{3}}{\frac{2}{3}} - 1 = -\frac{1}{2} - 1 = -\frac{1}{2} - \frac{2}{2} = -\frac{3}{2}
$$

Thus, the answers are:

$$
P(11) = -\frac{1}{12},\quad P\left(\frac{1}{8}\right) = -\frac{8}{9},\quad P\left(-\frac{1}{3}\right) = -\frac{3}{2}
$$
$$ P(11) = -\frac{1}{12},\quad P\left(\frac{1}{8}\right) = -\frac{8}{9},\quad P\left(-\frac{1}{3}\right) = -\frac{3}{2} $$

Let be the function . Find the following:

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