Question

choose any 5 problems

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 weather channel app gives the hourly temperature. here are select temperatures for a summer day in pittsburgh, pa. let $t = 0$ represent 6 am.

$\

$$\begin{array}{c|cccccc}t&0&3&6&9&12&15\\\\\\hline t(t)&60&73&85&87&84&75\\end{array}$$

$
how fast is temperature increasing, on average, between 9 am and noon?

1. let $h(t)$ give the height of a fast - growing tree, in feet, $t$ years after it is planted. which expression can be used to find the average rate at which the tree grows between the first and third year?

a) $\frac{h(3)-h(1)}{2}$
b) $\frac{h(1)-h(3)}{2}$
c) $\frac{h(1)+h(3)}{2}$
d) $\frac{2}{h(3)-h(1)}$

1. the graph of $y = h(x)$ is shown. find the average rate of change of $h$ on $-3,2$.

Explanation:

1.

Step1: Recall average rate - of - change formula

The average rate of change of a function $y = k(x)$ over the interval $[a,b]$ is $\frac{k(b)-k(a)}{b - a}$. Here, $a=-2$, $b = 4$, and $k(x)=-(x + 1)^2+11$.

Step2: Calculate $k(-2)$

Substitute $x=-2$ into $k(x)$:
\[

$$\begin{align*} k(-2)&=-(-2 + 1)^2+11\\ &=-(-1)^2+11\\ &=-1 + 11\\ &=10 \end{align*}$$

\]

Step3: Calculate $k(4)$

Substitute $x = 4$ into $k(x)$:
\[

$$\begin{align*} k(4)&=-(4 + 1)^2+11\\ &=-25+11\\ &=-14 \end{align*}$$

\]

Step4: Calculate the average rate of change

\[

$$\begin{align*} \frac{k(4)-k(-2)}{4-(-2)}&=\frac{-14 - 10}{4 + 2}\\ &=\frac{-24}{6}\\ &=-4 \end{align*}$$

\]

Step1: Identify the time - points

If $t = 0$ represents 6 am, then 9 am corresponds to $t=3$ and noon corresponds to $t = 6$.

Step2: Recall the average rate - of - change formula

The average rate of change of the temperature function $T(t)$ over the interval $[a,b]$ is $\frac{T(b)-T(a)}{b - a}$. Here, $a = 3$, $b = 6$, $T(3)=73$, and $T(6)=85$.

Step3: Calculate the average rate of change

\[

$$\begin{align*} \frac{T(6)-T(3)}{6 - 3}&=\frac{85-73}{3}\\ &=\frac{12}{3}\\ & = 4 \end{align*}$$

\]

The average rate of change of a function $y = h(t)$ over the interval $[a,b]$ is given by $\frac{h(b)-h(a)}{b - a}$. For the interval from the first year ($t = 1$) to the third year ($t = 3$), the average rate of change is $\frac{h(3)-h(1)}{3 - 1}=\frac{h(3)-h(1)}{2}$.

Answer:

$-4$

2.