Question

problem 1
the temperature was recorded at several times in a
24 - hour period. function ( f(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.
positive
negative
zero
( n = 1 ) to ( n = 5 )
positive
negative
zero
( n = 5 ) to ( n = 7 )
positive
negative
zero
( n = 10 ) to ( n = 20 )

Explanation:

Step1: Recall the formula for average rate of change

The average rate of change of a function \(y = f(x)\) over the interval \([a,b]\) is \(\frac{f(b)-f(a)}{b - a}\). In the context of the function \(t(n)\) (where \(y=t(n)\) and \(x = n\)), the average rate of change over the interval \([n_1,n_2]\) is \(\frac{t(n_2)-t(n_1)}{n_2 - n_1}\). Geometrically, if the function is increasing (going up from left - to - right) over the interval, the average rate of change is positive. If it is decreasing (going down from left - to - right), the average rate of change is negative.

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

Looking at the graph, as \(n\) (hours after midnight) increases from \(n = 1\) to \(n = 5\), the temperature \(t(n)\) (in degrees Fahrenheit) is increasing. For example, if we assume two points \((n_1,t(n_1))\) and \((n_2,t(n_2))\) with \(n_1 = 1\) and \(n_2=5\), since \(t(n_2)>t(n_1)\) and \(n_2 - n_1=5 - 1>0\), the average rate of change \(\frac{t(5)-t(1)}{5 - 1}>0\)

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

As \(n\) increases from \(n = 5\) to \(n = 7\), the temperature \(t(n)\) is decreasing. If we take two points \((n_1,t(n_1))\) with \(n_1 = 5\) and \((n_2,t(n_2))\) with \(n_2 = 7\), since \(t(n_2)<t(n_1)\) and \(n_2 - n_1=7 - 5>0\), the average rate of change \(\frac{t(7)-t(5)}{7 - 5}<0\)

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

As \(n\) increases from \(n = 10\) to \(n = 20\), the temperature \(t(n)\) is decreasing. If we take two points \((n_1,t(n_1))\) with \(n_1 = 10\) and \((n_2,t(n_2))\) with \(n_2 = 20\), since \(t(n_2)<t(n_1)\) and \(n_2 - n_1=20 - 10>0\), the average rate of change \(\frac{t(20)-t(10)}{20 - 10}<0\)

Answer:

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