What is the domain of \(g\)?

Evaluate \(g(-8) = \answer [tolerance=0.25]{-6}\)

Classify \(4\).

Evaluate \(g(-3) = \answer [tolerance=0.25]{5}\)

Classify \(-3\).

Evaluate \(g(4) = \answer [tolerance=0.25]{-2}\)

Classify \(4\).

Evaluate \(g(8) = \answer [tolerance=0.25]{8}\)

Classify \(4\).

\(g\) is increasing on \((-5, -3)\).

\(g\) is increasing on \([-5, -3]\).

\(g\) is decreasing on \((-3, 4)\).

\(g\) is decreasing on \([-3, 4]\).

There exists a real number, \(M\), such that \(g(t) < M\) for all \(t\).

There exists a real number, \(M\), such that \(g(t) > M\) for all \(t\).

There exists an \(\epsilon > 0\), such that for every \(d \in (0-\epsilon , 0+\epsilon )\) we have \(g(d) \in \left ( 2-\frac {1}{2}, 2+\frac {1}{2} \right )\).

There exists an \(\epsilon > 0\), such that for every \(d \in (0-\epsilon , 0+\epsilon )\) we have \(g(d) \in \left ( 2-\frac {1}{4}, 2+\frac {1}{4} \right )\).

There exists an \(\epsilon > 0\), such that for every \(d \in (0-\epsilon , 0+\epsilon )\) we have \(g(d) \in \left ( 2-\frac {1}{10}, 2+\frac {1}{10} \right )\).

For each \(N > 0\), there exists an \(\epsilon > 0\), such that for every \(d \in (0-\epsilon , 0+\epsilon )\) we have \(g(d) \in \left ( 2-\frac {1}{N}, 2+\frac {1}{N} \right )\).

There exists an \(\epsilon > 0\), such that for every \(d \in (4-\epsilon , 4+\epsilon )\) we have \(g(d) \in \left ( -2-\frac {1}{2}, -2+\frac {1}{2} \right )\).

There exists an \(\epsilon > 0\), such that for every \(d \in (-3-\epsilon , -3+\epsilon )\) we have \(g(d) \in \left ( 5-\frac {1}{4}, 5+\frac {1}{4} \right )\).