Question

find the average rate of change of the function graphed over the interval ( 10 leq x leq 50 ). compare it to the average rate of change of ( y = 4 log x + 18 ) over the same interval. the average rate of change of the graphed function is ( square ), and the average rate of change of ( y = 4 log x + 18 ) is ( square ), so the average rate of change of the graphed function is ( square ) the average rate of change of the given function. (type integers or decimals rounded to two decimal places as needed.)

Explanation:

Step1: Recall the formula for average rate of change

The formula for 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: Identify \(a\), \(b\), \(f(a)\) and \(f(b)\) for the graphed function

From the graph, when \(a = 10\), \(f(10)=19\) and when \(b = 50\), \(f(50)=27.5\).
Then \(\frac{f(50)-f(10)}{50 - 10}=\frac{27.5-19}{50 - 10}\).

Step3: Calculate the value for the graphed function

\(\frac{27.5-19}{50 - 10}=\frac{8.5}{40}=0.2125\approx0.21\).

Step4: Identify \(a\), \(b\), \(f(a)\) and \(f(b)\) for \(y = 4\log x+18\)

For \(y = 4\log x+18\), when \(x = 10\), \(y_1=4\log(10)+18=4\times1 + 18=22\). When \(x = 50\), \(y_2=4\log(50)+18\). Since \(\log(50)=\log(5\times10)=\log(5)+\log(10)\approx0.6990 + 1=1.6990\), then \(y_2=4\times1.6990+18=6.796+18 = 24.796\).

Step5: Calculate the average rate of change for \(y = 4\log x+18\)

\(\frac{y_2 - y_1}{50 - 10}=\frac{24.796-22}{40}=\frac{2.796}{40}=0.0699\approx0.07\).

Answer:

The average rate of change of the graphed function is \(0.21\), and the average rate of change of \(y = 4\log x+18\) is \(0.07\), so the average rate of change of the graphed function is greater than the average rate of change of \(y = 4\log x + 18\).