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 = 2 log x + 15 ) over the same
rate of change of the graphed function is (square), and the average rate of change of ( y = 2 log x + 15 ) is (square). so the average rate of change of the graphed function is (square) the average
(round answers as decimal rounded to two decimal places as needed.)

Explanation:

Step1: Recall the average rate of change formula

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: Calculate \(f(10)\) and \(f(50)\) for \(y = 2\log x+15\)

For \(x = 10\), \(y_1=2\log(10)+15\). Since \(\log(10) = 1\), \(y_1=2\times1 + 15=17\).
For \(x = 50\), \(y_2=2\log(50)+15\). Using \(\log(50)=\log(5\times10)=\log(5)+\log(10)\approx0.6990 + 1=1.6990\), then \(y_2=2\times1.6990+15=3.398+15 = 18.398\).

Step3: Apply the average rate of change formula

\(\frac{y_2 - y_1}{50 - 10}=\frac{18.398-17}{40}=\frac{1.398}{40}=0.03495\approx0.03\)

Answer:

The average rate of change of the graphed function is \(0.03\), and the average rate of change of \(y = 2\log x+15\) is \(0.03\). So the average rate of change of the graphed function is equal to the average rate of change of \(y = 2\log x + 15\) over the interval \(10\leq x\leq50\).