f0eff4…: “Far from the useful lake”

$\newenvironment {prompt}{}{} \newcommand {\accordion }[0]{} \newcommand {\ungraded }[0]{} \newcommand {\xmDefaultPreamble }[0]{xmPreamble.tex} \newcommand {\xmDefaultPrintstyle }[0]{xmPrintstyle.sty} \newcommand {\todo }[0]{} \newcommand {\oiint }[0]{{\large \bigcirc }\kern -1.56em\iint } \newcommand {\mooculus }[0]{\textsf {\textbf {MOOC}\textnormal {\textsf {ULUS}}}} \newcommand {\sagemath }[0]{\textsf {SageMath}} \newcommand {\surfaceColor }[0]{violet} \newcommand {\surfaceColorTwo }[0]{redyellow} \newcommand {\sliceColor }[0]{greenyellow} \newcommand {\vector }[1]{\left \langle #1\right \rangle } \newcommand {\sectionOutcomes }[0]{}$

Define $V$ as follows.

\[ V(h) = \begin{cases} -2h-3 & \text { if } [-6, -2] \\ -(h+3)(h-3) & \text { if } (-2, 4] \\ \frac {7h}{4} - 8 & \text { if } (4,6) \end{cases} \]

$V$ has many features, characteristics, and aspects.

The domain of $V$ is

\[ \left [ \answer {-6}, \answer {6} \right ) \]

Let $g(t) = 4 V(2t - 5) - 1$.

The domain of $g$ is

\[ \left [ \answer {\frac {-1}{2}}, \answer {\frac {11}{2}} \right ) \]

Let $H(b) = -5 V(7 - 3b) + 2$.

The domain of $H$ is

\[ \left ( \answer {\frac {1}{3}}, \answer {\frac {13}{3}} \right ] \]

2025-05-18 11:42:33