<h3>Answer: c = pi/2</h3>
There is only one value of c that satisfies the requirements.
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Work Shown:
Let's compute the function value at the given endpoints of the interval.
![f(x) = 4\sin(x)\\\\f(\pi) = 4\sin(\pi) = 0\\\\f(0) = 4\sin(0) = 0\\\\](https://tex.z-dn.net/?f=f%28x%29%20%3D%204%5Csin%28x%29%5C%5C%5C%5Cf%28%5Cpi%29%20%3D%204%5Csin%28%5Cpi%29%20%3D%200%5C%5C%5C%5Cf%280%29%20%3D%204%5Csin%280%29%20%3D%200%5C%5C%5C%5C)
Which means,
![f'(c) = \frac{f(b)-f(a)}{b-a}\\\\f'(c) = \frac{f(\pi)-f(0)}{\pi-0}\\\\f'(c) = \frac{0-0}{\pi}\\\\f'(c) = \frac{0}{\pi}\\\\f'(c) = 0\\\\](https://tex.z-dn.net/?f=f%27%28c%29%20%3D%20%5Cfrac%7Bf%28b%29-f%28a%29%7D%7Bb-a%7D%5C%5C%5C%5Cf%27%28c%29%20%3D%20%5Cfrac%7Bf%28%5Cpi%29-f%280%29%7D%7B%5Cpi-0%7D%5C%5C%5C%5Cf%27%28c%29%20%3D%20%5Cfrac%7B0-0%7D%7B%5Cpi%7D%5C%5C%5C%5Cf%27%28c%29%20%3D%20%5Cfrac%7B0%7D%7B%5Cpi%7D%5C%5C%5C%5Cf%27%28c%29%20%3D%200%5C%5C%5C%5C)
We want to find all values of c such that the derivative is 0.
This can be rephrased into wanting to solve
since the derivative of sine is cosine
In other words,
turns into ![f'(x) = 4\cos(x)\\\\](https://tex.z-dn.net/?f=f%27%28x%29%20%3D%204%5Ccos%28x%29%5C%5C%5C%5C)
Solving that equation leads to:
![4\cos(x) = 0\\\\\cos(x) = 0\\\\x = \frac{\pi}{2}\\\\](https://tex.z-dn.net/?f=4%5Ccos%28x%29%20%3D%200%5C%5C%5C%5C%5Ccos%28x%29%20%3D%200%5C%5C%5C%5Cx%20%3D%20%5Cfrac%7B%5Cpi%7D%7B2%7D%5C%5C%5C%5C)
Use a reference table or the unit circle to determine this.
This is the value of c that satisfies f ' (c) = 0, such that
. No other value of c works.
If you graphed f(x) = 4sin(x), and only focused on the interval [0,pi], then you'll find that there's a horizontal tangent at the point (pi/2, 4). Note how the endpoints have the same y value so that's why the average rate of change over [0,pi] is 0.
Side note: this is an application of Rolle's Theorem