How do you determine all values of c that satisfy the mean value theorem on the interval [pi/2, 3pi/2] for # f(x) = sin(x/2)#?

Question

34020 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-04T18:01:12

The value is #c=pi#

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

Therefore there exists a #c# on #( pi/2, 3pi/2)# such that

#f'(c) = ( f(3pi/2) - f(pi/2) ) / ( 3pi/2 -pi/2)=>f'(c)=(f(3pi/2)-f(pi/2))/pi#

We know that #f(x) = sin(x/2)# hence
#f(3pi/2) = sin (3pi/4) = sqrt2/2#
#f(pi/2) = sin(pi/4) = sqrt2/2#

We notice that

#( f(3pi/2) - f(pi/2) ) / pi = (sqrt2/2 - sqrt2/2) / pi = 0#

But

#f'(x) = (1/2) cos(x/2)#

#f'(c) = (1/2) cos(c/2) = 0=>cos(c/2)=0=>c/2=pi/2=>c=pi#