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If, f(x)=sin(x/2), then there exists a number c in the interval  pi/2 < x < 3pi/2  that satisfies the conclusion of the Mean Value Theorem. What is the value of c ?

Apr 9, 2021
edited by wick3  Apr 9, 2021

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The mean value theorem here says:

Can you finish from here?

Apr 9, 2021
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If, f(x)=sin(x/2), then there exists a number c in the interval  pi/2 < x < 3pi/2  that satisfies the conclusion of the Mean Value Theorem. What is the value of c ?

Hello Guest!

$$f(x)=sin(\frac{x}{2})\\ f(\frac{\pi}{2})=sin(\frac{\pi}{4})=\frac{1}{2}\sqrt{2}\\ f(\frac{3\pi}{2})=sin(\frac{3\pi}{4})=\frac{1}{2}\sqrt{2}$$

The secant through these function values has a slope of 0, , like the tangent in the maximum of the function.

$$f(x)=sin(\frac{x}{2})\\ \frac{df(x)}{dx}=\frac{1}{2}cos(\frac{x}{2})=0\\ \color{blue}x_{max}=\pi$$

$$f(x)=sin(\frac{x}{2})\\ \color{blue}f_{max}(x)=f(\pi)=sin(\frac{\pi}{2})=1$$

$$\color{blue}The\ value\ of\ c\ is\ (\pi,1).$$

!

Apr 9, 2021