Solutions

a) We can assert that $f$ has a global minimum since it is a polynomial of even degree with a positive leading coefficient. A thorough analysis of why this is true first comes from realizing that

$$\lim_{x \to \infty} f(x) = \lim_{x \to -\infty} f(x) = \infty.$$

This means that for any $M > 0$, there exists $\delta > 0$ such that

$$\forall x \in \mathbb{R}, \quad |x| > \delta \implies f(x) > M.$$

Let $M = f(0)$. Then,

$$\forall x \in \mathbb{R}, \quad |x| > \delta \implies f(x) > f(0).$$

Now consider the interval $[-M, M]$. We know that $f$ is continuous over this interval (since $f$ is a sum of continuous functions) and therefore it attains a minimum value at $f(c)$ for some point $c \in [-M, M]$. Hence,

$$\forall x \in [-M, M], \quad f(c) \le f(x)$$

and

$$\forall x \notin [-M, M], \quad f(c) \le f(0) < f(x).$$

Hence, $f(c) \le f(x)$ for all $x$ and therefore $c$ is the global minimum point of $f$.

We can now apply the interior extremum theorem to the interval $(-M, M)$ over which $c$ is a minimum point and at which $f$ is also differentiable, being a polynomial, and get

$$\begin{align*} f'(c) = 0 &\implies \sum_{i=1}^n 2(c - a_i) = 0 \\ &\implies nc = \sum_{i=1}^n a_i \\ &\implies c = \frac{\sum_{i=1}^n a_i}{n}. \end{align*}$$

b) Spivak gives us the hint of considering $f(y) - f(x)$ for $x \in [a_{j-1}, a_j]$ and $y \in [a_j, a_{j+1}]$. The expression gives

$$\begin{align*} f(y) - f(x) &= \sum_{i=1}^n |y - a_i| - \sum_{i=1}^n |x - a_i| \\ &= \sum_{i=1}^n \left(|y - a_i| - |x - a_i|\right). \end{align*}$$

Notice that for $i \ge j + 1$,

$$|y - a_i| - |x - a_i| = -(y - x)$$

and for $i \le j - 1$,

$$|y - a_i| - |x - a_i| = y - x.$$

Thus,

$$f(y) - f(x) = (j - 1)(y - x) - (n - j)(y - x) + |x - a_j| - |y - a_j|$$

and therefore

$$f(y) - f(x) = (2j - n - 1)(y - x) + |x - a_j| - |y - a_j|.$$

Now if $|x - a_j| = |y - a_j|$, then

$$f(y) - f(x) = (2j - n - 1)(y - x).$$

Notice that for odd values of $n$, the difference is negative for $j < \frac{n+1}{2}$, $0$ at the midpoint, and positive for $j > \frac{n+1}{2}$. In this case the minimum value is

$$f\left(a_{\frac{n+1}{2}}\right).$$

For even values of $n$, we notice the same decreasing and increasing function pattern but over $j \le \frac{n}{2}$ and $j > \frac{n}{2}$ respectively. This hints at the minimum being within the interval $\left[a_{\frac{n}{2}}, a_{\frac{n}{2} + 1}\right]$. In fact, if $x, y \in \left[a_{\frac{n}{2}}, a_{\frac{n}{2} + 1}\right]$ and $x \le y$ then

$$|y - a_j| - |x - a_j| = \begin{cases} y - x, & j \le \frac{n}{2}, \\ -(y - x), & j > \frac{n}{2}. \end{cases}$$

Hence,

$$\begin{align*} f(y) - f(x) &= \sum_{i=1}^n |y - a_i| - |x - a_i| \\ &= \frac{n}{2}(y - x) + \frac{n}{2} (-(y - x)) \\ &= 0. \end{align*}$$

Thus, for even values of $n$ we conclude that the minimum value is $f(z)$ for $z \in \left[a_{\frac{n}{2}}, a_{\frac{n}{2} + 1}\right]$.

c) We have that

$$f'(x) = -\frac{\operatorname{sign}(x)}{(1 + |x|)^2} - \frac{\operatorname{sign}(x - a)}{(1 + |x - a|)^2}$$

for $x \ne 0$ and $x \ne a$. Consider now its critical points:

$$f'(x) = 0 \implies \frac{\operatorname{sign}(x)}{(1 + |x|)^2} = -\frac{\operatorname{sign}(x - a)}{(1 + |x - a|)^2}.$$

We see that $f$ has critical points neither for $x < 0$ nor for $x > a$. For $0 < x < a$, we have

$$\begin{align*} f'(x) = 0 &\implies \frac{1}{(1 + |x|)^2} = \frac{1}{(1 + |x - a|)^2} \\ &\implies |x| = |x - a| \\ &\implies x^2 = (x - a)^2 \\ &\implies x^2 = x^2 - 2ax + a^2 \\ &\implies x = \frac{a}{2}. \end{align*}$$

The values of $f$ at notable points include

$$f(0) = 1 + \frac{1}{1 + a},$$$$f\left(\frac{a}{2}\right) = \frac{1}{1 + \frac{a}{2}} + \frac{1}{1 + \frac{a}{2}} = \frac{4}{2 + a},$$$$f(a) = \frac{1}{1 + a} + 1.$$

Since $f'$ is positive for $x < 0$, negative for $0 < x < \frac{a}{2}$, positive again for $\frac{a}{2} < x < a$, and negative again for $x > a$, then the maximum value of $f$ is

$$\frac{1}{1 + a} + 1.$$