MHB Determine the Average Rate of Change

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The average rate of change (AROC) of the function y = 2cos(x - π/3) + 1 over the interval π/3 ≤ x ≤ π/2 was initially calculated incorrectly, yielding an approximate value of 1.4. Upon further analysis, the correct AROC was determined to be approximately -0.5117 after identifying errors in the calculations. The discrepancy was clarified by graphing the function and analyzing the secant line. This highlights the importance of careful computation and verification in determining the AROC. The final correct value reflects the function's behavior over the specified interval.
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Determine the average rate if change of the function y = 2cos(x - $\pi$/3) + 1 for the interval $\pi$/3 $\le$ x $\le$ $\pi$/2

I tried finding the exact values of the two (0 and 0.5) and subbing them into the AROC equation but I keep getting the wrong answer (1.4)
 
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We find:

$$\frac{\Delta y}{\Delta x}=\frac{y\left(\dfrac{\pi}{2}\right)-y\left(\dfrac{\pi}{3}\right)}{\dfrac{\pi}{2}-\dfrac{\pi}{3}}=\frac{\left(2\cos\left(\dfrac{\pi}{6}\right)+1\right)-\left(2\cos\left(0\right)+1\right)}{\dfrac{\pi}{6}}=\frac{6\left(\sqrt{3}-1\right)}{\pi}\approx1.39811405542801$$
 
MarkFL said:
We find:

$$\frac{\Delta y}{\Delta x}=\frac{y\left(\dfrac{\pi}{2}\right)-y\left(\dfrac{\pi}{3}\right)}{\dfrac{\pi}{2}-\dfrac{\pi}{3}}=\frac{\left(2\cos\left(\dfrac{\pi}{6}\right)+1\right)-\left(2\cos\left(0\right)+1\right)}{\dfrac{\pi}{6}}=\frac{6\left(\sqrt{3}-1\right)}{\pi}\approx1.39811405542801$$

Okay, that's what I got. The textbook has it as -0.5157 so I guess it's just wrong?
 
We both made an error...it should be:

$$\frac{\Delta y}{\Delta x}=\frac{6\left(\sqrt{3}-2\right)}{\pi}\approx-0.511745261674736$$

I discovered my error when graphing the function and the resulting secant line:

View attachment 3547
 

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