+15001 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.
\(f(x)=sin(\frac{x}{2})\\ \frac{df(x)}{dx}=\frac{1}{2}cos(\frac{x}{2})=0\\ \color{blue}x_{max}=\pi\)
\(f(x)=sin(\frac{x}{2})\\ \color{blue}f_{max}(x)=f(\pi)=sin(\frac{\pi}{2})=1\)
\(\color{blue}The\ value\ of\ c\ is\ (\pi,1).\)
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