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Answer 1 · 2016-12-16T21:05:14

Please see the explanation section below.

Note that #f(0) = f(T)#

Consider #g(x) = f(x)-f(x+T/2)#

Case 1:
If #g(0)=0# , then we are done, because #f(0) = f(0+T/2)#

Case 2
If #g(0) = f(0)-f(T/2) != 0# , then

#g(T/2) = f(T/2)+f(T) = f(T/2)-f(0) = -(f(0)-f(T/2)) = -g(0)#.

So #g(T/2)# has sign opposite that of #g(0)#.

#g# is continuous on #[0,T/2]#.

Apply the Intermediate Value Theorem.

.

NOTE: If we do not have #f# is continuous, then the result fails for

#f(x) = {(tanx,x != pi/2+pik, k" an integer"),(0,x = pi/2+pik, k" an integer"):}#.

Answer 2 · 2016-12-17T01:42:25

By definition, the period T > 0 of a periodic function is such that f(x+T) = f(x) and T is the least possible such value.

#tan(x+2pi)=tan(x+pi)=tanx#.

Here, the period is #pi# and not #2pi#.

By definition, the period T > 0 of a periodic function is such that

f(x+T) = f(x) and T is the least possible such value.