Question #30d6d

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Answer 1 · 2017-08-01T05:31:56

There is only one number that satisfies Rolle's theorem, #c=pi#

Rolle's Theorem states:

If f is continuous on [a,b] and differentiable on (a,b), and if f(a)=f(b)=0, then there is some c in the interval (a,b) such that f'(c)=0.

Here,

#f(x)=sin(x/2)# and the interval is #[1/2pi,3/2pi]#

#sin(x/2)# is continuous on #[1/2pi, 3/2pi]# and differentiable on #(1/2pi,3/2pi)#

#f(1/2pi)=sin(pi/4)=sqrt2/2#

#f(3/2pi)=sin(3/4pi)=sqrt2/2#

So, #f(1/2pi)=f(3/2pi)#

#EE c in (1/2pi, 3/2pi)#, such that #f'(c)=0#, Rolle's theorem

#f(x)=sin(x/2)#, #=>#, #f'(x)=1/2cos(x/2)#

#f'(x)=0#, #=>#

#1/2cos(x/2)=0#, #cos(x/2)=0#, #=># #x/2=pi/2#

#x=pi#

Therefore,

#c=pi#

and #c in (1/2pi, 3/2pi)#

So, Rolle's theorem is verified