Question

Question #4cbca

Answer 1

Please see the explanation section below.

Explanation:

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

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.

Explanation:

`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.