Completely analyze \(G(t) = -2 | t - 3 | + 6\) with its natural domain.

Domain:

\(G\) is an absolute value function, therefore its natural domain is \((-\infty , \infty )\).

Continuity:

\(G\) is an absolute value function, therefore it is continuous on its domain and has no discontinuities.

\(G\) is an absolute value function, therefore it has no singularities.

End-Behavior:

\(\blacktriangleright \) As \(t\) becomes big, then the inside becomes big positively or negatively. Either way, the \(| t - 3|\) is big and positive. Since the leading coefficient is negative, we have

\[ \lim \limits _{t \to \infty } G(t) = -\infty \]
\[ \lim \limits _{t \to -\infty } G(t) = -\infty \]

Behavior:

\(G\) is an absolute value function, therefore, it is a piecewise function. The pieces are linear functions.

\[ G(t) = \begin{cases} -2 ( t - 3 ) + 6 & (-\infty , 5)\\ -2 (3 - t) + 6 & (5, \infty ) \end{cases} \]
\[ G(t) = \begin{cases} -2 t + 12 & (-\infty , 5)\\ 2t & (5, \infty ) \end{cases} \]

\(-2 t + 12\) is a decreasing linear function. \(G(t)\) decreases on \((-\infty , 5)\).

\(2 t\) is an increasing linear function. \(G(t)\) increases on \((5, \infty )\).

Zeros:

We can find the zeros of each piece of the piecewise function.

\(-2 t + 12 \) has a zero at \(6\).

\(-2 t\) has a zero at \(0\).

Extema:

From the end-behavior, we know that \(G\) has no global or local minimum.

Since \(G\) is an absolute value funciton, \(G\) has a global and local maximum when the inside is \(0\), which is at \(3\).

\(G\) is a global and local maximum of \(G(3) + 6\) at \(3\).

This all agrees with the graph.

2025-01-07 02:00:06