Question

1. let $k(x)=-(x + 1)^2+11$. find the average rate of change of $k(x)$ between $x=-2$ and $x = 4$.
2. the temperature of a mug of hot chocolate, $t$ minutes after it is prepared is given by the function $h(t)$ where $h$ is measured in degrees fahrenheit. selected values of $h(t)$ are given below.

$\

$$\begin{array}{c|cccccc}t&0&2&5&10&13&20\\hline h(t)&170&137&107&83&77&72\\end{array}$$

$
find the average rate of change of the temperature of the hot chocolate between $t = 0$ and $t=20$. use proper units in your answer.

Explanation:

Step1: Recall average - rate - of - change formula

The average rate of change of a function $y = f(x)$ over the interval $[a,b]$ is $\frac{f(b)-f(a)}{b - a}$.

Step2: For $k(x)=-(x + 1)^2+11$, find $k(-2)$ and $k(4)$

When $x=-2$, $k(-2)=-(-2 + 1)^2+11=-(-1)^2+11=-1 + 11 = 10$.
When $x = 4$, $k(4)=-(4 + 1)^2+11=-25+11=-14$.

Step3: Calculate average rate of change of $k(x)$

The average rate of change of $k(x)$ over $[-2,4]$ is $\frac{k(4)-k(-2)}{4-(-2)}=\frac{-14 - 10}{4 + 2}=\frac{-24}{6}=-4$.

Step4: For the temperature function $H(t)$

We know $a = 0$, $H(0)=170$, $b = 20$, $H(20)=72$.

Step5: Calculate average rate of change of $H(t)$

The average rate of change of $H(t)$ over $[0,20]$ is $\frac{H(20)-H(0)}{20-0}=\frac{72-170}{20}=\frac{-98}{20}=-4.9$ degrees Fahrenheit per minute.

Answer:

1. The average rate of change of $k(x)$ between $x=-2$ and $x = 4$ is $-4$.
2. The average rate of change of the temperature of the hot chocolate between $t = 0$ and $t=20$ is $-4.9$ degrees Fahrenheit per minute.