Find minimum of an expression $f(q) = log_2(q^q \cdot (n-q+1)^{n-q+1})$

Question

I have that following expression:

$$ f(q) = log_2(q^q \cdot (n-q+1)^{n-q+1}) $$

The solution says it has a min at

$$ q = (n+1)/2 $$

Is there a strait way to get to this min?

When i try to take the derivate it becomes complicated (or i miss something)?

Anyway, how did they get to this min?

Answer 1

$$f(q)=\log_2(q^q \cdot (n-q+1)^{n-q+1} )\\ = q\log_2 q + (n-q+1)\log_2 (n-q+1)$$

$$f’(q) = q\cdot \frac 1q + \log _2 (q) \frac{1}{\ln 2} - (n-q+1) \cdot \frac{1}{n-q+1} - \log_2(n-q+1) \frac{1}{\ln 2}=0 \\ \implies \log_2(q) = \log_2(n-q+1) \\ \implies q=\frac{n+1}{2}$$