fef88b…: “Upon the interesting tree”

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

Let $H(x)$ be a quadratic function with $iRoC_H(x) = 5x - 10$.

$iRoC_H(x) = 0$ at $t = \answer {2}$.

$iRoC_H(x) < 0$ on $\left ( \answer {-\infty }, \answer {2} \right )$.

$iRoC_H(x) > 0$ on $\left ( \answer {2}, \answer {\infty } \right )$.

$H(x)$ decreases on $\left ( \answer {-\infty }, \answer {2} \right )$.

$H(x)$ increases on $\left ( \answer {2}, \answer {\infty } \right )$.

The minimum of $H(x)$ is located at $x = \answer {2}$.

2025-05-18 05:22:13