+0  
 
0
135
2
avatar+8 

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
 #1
avatar+32123 
+3

The mean value theorem here says:

 

Can you finish from here?

 Apr 9, 2021
 #2
avatar+11418 
+1

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.

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

 

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

 

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

laugh  !

 Apr 9, 2021

67 Online Users

avatar
avatar
avatar
avatar
avatar
avatar