Asked by Anonymous
                Find the POSITIVE value of x that satisfies the mean value theorem for 
f(x)=sin(x) on the closed interval
[-3pi/2, 3pi/2].
please help I have no idea how to solve this problem
            
        f(x)=sin(x) on the closed interval
[-3pi/2, 3pi/2].
please help I have no idea how to solve this problem
Answers
                    Answered by
            Damon
            
    sin 3 pi/2 = sin 270 = -1
sin - 3 pi/2 = +1
change in sin x = -1 - +1 = -2
change in x = 3 pi/2+3pi/2 = 3 pi
so
slope from left to right = -2 / 3pi
= -.2122
so where does the derivative of sin x = -.2122 ?
d/dx (sin x) = cos x
so cos x = -.2122
x = 102.25 degrees = 1.78 radians
    
sin - 3 pi/2 = +1
change in sin x = -1 - +1 = -2
change in x = 3 pi/2+3pi/2 = 3 pi
so
slope from left to right = -2 / 3pi
= -.2122
so where does the derivative of sin x = -.2122 ?
d/dx (sin x) = cos x
so cos x = -.2122
x = 102.25 degrees = 1.78 radians
                                                    There are no AI answers yet. The ability to request AI answers is coming soon!
                                            
                Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.