Question

1. given the function f(x)=3x² - 2x + 15 and the function g(x) shown in the table below, which has the greater average rate of change, f(x) on the interval 3 ≤ x ≤ 7 or g(x) on the interval 4 ≤ x ≤ 9? show calculations to support your answer.

x	0	1	4	6	9	10
g(x)	0	24	82	182	257	325

1. the temperature of a greenhouse in degrees fahrenheit is shown graphically as a function of the time in hours since midnight.

(a) what is the average rate of change of the temperature from 2am until 12pm (noon)?
(b) is the temperature decreasing faster on the interval 0 ≤ t ≤ 2 or 18 ≤ t ≤ 22? show the analysis that justifies your conclusion.

1. as we have seen, the average rate of change of most functions depends on what domain interval is chosen. there is one type of function that has a constant average rate of change (meaning it does not depend on what domain interval is chosen.). what type of function is this? do an internet search if you need to.

Explanation:

Step1: Recall average - rate - of - change formula

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

Step2: Calculate average rate of change of $f(x)=3x^{2}-2x + 15$ on $[3,7]$

First, find $f(3)$ and $f(7)$.
$f(3)=3\times3^{2}-2\times3 + 15=3\times9-6 + 15=27-6 + 15=36$.
$f(7)=3\times7^{2}-2\times7 + 15=3\times49-14 + 15=147-14 + 15=148$.
The average rate of change of $f(x)$ on $[3,7]$ is $\frac{f(7)-f(3)}{7 - 3}=\frac{148 - 36}{4}=\frac{112}{4}=28$.

Step3: Calculate average rate of change of $g(x)$ on $[4,9]$

From the table, $g(4)=82$ and $g(9)=257$.
The average rate of change of $g(x)$ on $[4,9]$ is $\frac{g(9)-g(4)}{9 - 4}=\frac{257-82}{5}=\frac{175}{5}=35$.
Since $35>28$, $g(x)$ on the interval $4\leq x\leq9$ has a greater average rate of change.

Step4: For problem 7(a)

From the graph, at 2 AM ($t = 2$), the temperature $F(2)\approx60$. At 12 PM ($t = 12$), the temperature $F(12)\approx75$.
The average rate of change of the temperature from 2 AM until 12 PM is $\frac{F(12)-F(2)}{12 - 2}=\frac{75 - 60}{10}=1.5$.

Step5: For problem 7(b)

For the interval $0\leq t\leq2$:
Let $F(0)\approx67$ and $F(2)\approx60$. The average rate of change is $\frac{F(2)-F(0)}{2-0}=\frac{60 - 67}{2}=-3.5$.
For the interval $18\leq t\leq22$:
Let $F(18)\approx77$ and $F(22)\approx68$. The average rate of change is $\frac{F(22)-F(18)}{22 - 18}=\frac{68 - 77}{4}=-2.25$.
Since $\vert-3.5\vert>\vert-2.25\vert$, the temperature is decreasing faster on the interval $0\leq t\leq2$.

Step6: For problem 8

A linear function $y=mx + b$ has a constant average rate of change (the slope $m$). The average rate of change of a linear function $y = mx + b$ on any interval $[a,b]$ is $\frac{(mb + b)-(ma + b)}{b - a}=\frac{m(b - a)}{b - a}=m$.

Answer:

1. $g(x)$ on the interval $4\leq x\leq9$ has a greater average rate of change.
2. (a) The average rate of change of the temperature from 2 AM until 12 PM is $1.5$.

(b) The temperature is decreasing faster on the interval $0\leq t\leq2$.

1. A linear function has a constant average rate of change.