Question

problem 1
the temperature was recorded at several times in a 24 - hour period. function ( t(n) ) gives the temperature in degrees fahrenheit ( n ) hours after midnight.
use the graph to determine if the average rate of change for each interval is positive, negative, or zero.
( n = 1 ) to ( n = 5 )
positive negative zero
( n = 5 ) to ( n = 7 )
positive negative zero
( n = 10 ) to ( n = 20 )
positive negative zero

Explanation:

Step1: Understand average rate of change concept

The average rate of change of a function \(y = f(x)\) over the interval \([x_1,x_2]\) is \(\frac{f(x_2)-f(x_1)}{x_2 - x_1}\). If \(f(x_2)>f(x_1)\) (when \(x_2>x_1\)), the average rate of change is positive. If \(f(x_2)<f(x_1)\) (when \(x_2>x_1\)), the average rate of change is negative.

Step2: Analyze \(n = 1\) to \(n = 5\)

Looking at the graph, when \(n = 1\), the temperature \(t(1)\approx42\) and when \(n = 5\), \(t(5)\approx36\). Since \(t(5)<t(1)\), the average rate of change \(\frac{t(5)-t(1)}{5 - 1}=\frac{36 - 42}{4}=\frac{-6}{4}<0\)

Step3: Analyze \(n = 5\) to \(n = 7\)

When \(n = 5\), \(t(5)\approx36\) and when \(n = 7\), \(t(7)\approx42\). Since \(t(7)>t(5)\), the average rate of change \(\frac{t(7)-t(5)}{7 - 5}=\frac{42-36}{2}=\frac{6}{2}>0\)

Step4: Analyze \(n = 10\) to \(n = 20\)

When \(n = 10\), \(t(10)\approx68\) and when \(n = 20\), \(t(20)\approx64\). Since \(t(20)<t(10)\), the average rate of change \(\frac{t(20)-t(10)}{20 - 10}=\frac{64 - 68}{10}=\frac{-4}{10}<0\)

Answer:

\(n = 1\) to \(n = 5\): Negative; \(n = 5\) to \(n = 7\): Positive; \(n = 10\) to \(n = 20\): Negative