Answer:
Step-by-step explanation:
The average value theorem sets:
if f (x) is continuous in [a, b] and derivable in (a, b) there is a c Є (a, b) such that
[tex]\frac{f(b)-f(a)}{b-a}=f'(c)[/tex] , where
f(a)=f(π/2)=-4*sin(π/2) = -4*1= -4
f(b)=(3π/2)=-4*sin(3π/2) = -4*-1 = 4
[tex]\frac{4-(-4)}{(3\pi/2)-(\pi/2)}=f'(c)[/tex]
[tex]\frac{8}{\pi }=f'(c)[/tex]
[tex]f'(x)=-4cos(x)[/tex] ⇒
[tex]f'(c)=-4cos(c)=\frac{8}{\pi }\\c=acos(\frac{-2}{\pi })\\[/tex]
c≅130
