# Question #4cbca

Dec 16, 2016

Please see the explanation section below.

#### Explanation:

Note that $f \left(0\right) = f \left(T\right)$

Consider $g \left(x\right) = f \left(x\right) - f \left(x + \frac{T}{2}\right)$

Case 1:
If $g \left(0\right) = 0$ , then we are done, because $f \left(0\right) = f \left(0 + \frac{T}{2}\right)$

Case 2
If $g \left(0\right) = f \left(0\right) - f \left(\frac{T}{2}\right) \ne 0$ , then

$g \left(\frac{T}{2}\right) = f \left(\frac{T}{2}\right) + f \left(T\right) = f \left(\frac{T}{2}\right) - f \left(0\right) = - \left(f \left(0\right) - f \left(\frac{T}{2}\right)\right) = - g \left(0\right)$.

So $g \left(\frac{T}{2}\right)$ has sign opposite that of $g \left(0\right)$.

$g$ is continuous on $\left[0 , \frac{T}{2}\right]$.

Apply the Intermediate Value Theorem.

.

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

$f \left(x\right) = \left\{\begin{matrix}\tan x & x \ne \frac{\pi}{2} + \pi k & k \text{ an integer" \\ 0 & x = pi/2+pik & k" an integer}\end{matrix}\right.$.

Dec 17, 2016

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:

$\tan \left(x + 2 \pi\right) = \tan \left(x + \pi\right) = \tan x$.

Here, the period is $\pi$ and not $2 \pi$.

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.