Answer: c = pi/2
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.
[tex]f(x) = 4\sin(x)\\\\f(\pi) = 4\sin(\pi) = 0\\\\f(0) = 4\sin(0) = 0\\\\[/tex]
Which means,
[tex]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\\\\[/tex]
We want to find all values of c such that the derivative is 0.
This can be rephrased into wanting to solve [tex]4\cos(x) = 0\\\\[/tex] since the derivative of sine is cosine
In other words, [tex]f(x) = 4\sin(x)\\\\[/tex] turns into [tex]f'(x) = 4\cos(x)\\\\[/tex]
Solving that equation leads to:
[tex]4\cos(x) = 0\\\\\cos(x) = 0\\\\x = \frac{\pi}{2}\\\\[/tex]
Use a reference table or the unit circle to determine this.
This is the value of c that satisfies f ' (c) = 0, such that [tex]0 < c < \pi[/tex]. 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
