Exercise 7.4.3

Show that the expression $q^2 + (n - q - 1)^2$ achieves a maximum over $q = 0, 1, \ldots, n-1$ when $q = 0$ and $q = n - 1$.

$$ \begin{align} f(q) &= q^2 + (n - q - 1)^2 \\ f'(q) &= 2q - 2(n - q - 1) = 4q - 2n + 2 \\ f''(q) &= 4 \\ \end{align} $$

$f'(q) = 0$ when $q = \frac{1}{2}n - \frac{1}{4}$. $f'(q)$ is also continious. $\forall q : f''(q) > 0$, which means that $f'(q)$ is negative left of $f'(q) = 0$ and positive right of it, which means that this is a local minima. In this case, $f(q)$ is decreasing in the beginning of the interval and increasing in the end, which means that those two points are the only candidates for a maximum in the interval.

$$ f(0) = (n - 1)^2 \\ f(n-1) = (n - 1)^2 + 0^2 $$