Question involving concept of $\max\{.\}\min\{.\}$

Question

In a book I’m been reading and solving from, I encountered a problem that is troubling me for too long as I can’t understand the concept behind the $\max, \min$ calls in the question and have been wondering if it’s a typographical error but either way I’d want to know for sure.

Let $f(x)=f_1(x) - 2f_2(x)$

where $f_1(x)= \begin{cases}\min\{x^2,|x|\}, & \text{$|x|\le1$} \\ \max\{x^2,|x|\},& \text{$|x| > 1$} \end{cases}$

Therefore, $ f_1(x) = x^2$, $x\in R$

and $ f_2(x) = \begin{cases}\min\{x^2,|x|\}, & \text{$|x|>1$} \\ \max\{x^2,|x|\},& \text{$|x|\le 1$} \end{cases}$

Therefore, $f_2(x) = |x|, x\in R$

Till here I understood the max, min calls. Now, the next function is what I’m not able to understand-

and let $g(x) = \begin{cases}\min\{f(t):-3 \le t \le x,& \text{$-3 \le x<0$}\} \\ \max\{f(t): 0 \le t \le x,& \text{$0\le x \le 3$}\} \end{cases}$

So what exactly does $g(x)$ mean? Does the above function $g(x)$ actually signify something or is there something missing in the space? I checked the solution as I couldn’t go further in the question without solving this doubt. In the solution provided, it says

$g(x) =\begin{cases} f(x), & \text{$-3 \le x < -1$} \\ -1, & \text{$-1 \le x < 0$}\\ 0, & \text{$0\le x \le 2$}\\ f(x), & \text{$2 < x \le 3$}\\ \end{cases}$

$g(x) =\begin{cases} x^2 + 2x, & \text{$-3 \le x < -1$} \\ -1, & \text{$-1 \le x < 0$}\\ \tag{1} 0, & \text{$0\le x \le 2$}\\ x^2 - 2x, & \text{$2 < x \le 3$}\\ \end{cases}$

How to handle the $t$ along with the $x$ and prove $g(x) = (1)$? Or is there something missing in the definition of $g(x)$ that makes $g(x) = (1)$? Any help would be appreciated.

Answer 1

You have concluded that $f_1(x)=x^2$ and $f_2(x)=|x|$.

Given, $f(x)=f_1(x)-2f_2(x)$

When, $x>0$

$$f(x)=x^2-2x$$

When $x<0$

$$f(x)=x^2+2x$$

You can easily calculate that $f(x)$ achieves its local mininum $x=+1,-1$ as $f(x)=-1$ and local maximum at $x=0$ as $f(x)=0$.

You are given that, $g(x) = \begin{cases}\min\{f(t):-3 \le t \le x,& \text{$-3 \le x<0$}\} \\ \max\{f(t): 0 \le t \le x,& \text{$0\le x \le 3$}\} \end{cases}$

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CASE I

As long as $x\lt-1$.

$f(x)$ i.e. the closing element of the set will correspond to the functions minimum value.

But as $x\geq-1$

There will always lie a $t=-1$ in $-3\leq{t}\leq{x}$ such that $f(t)$ can achieve its minimum value, i.e. $f(t)=-1$ at $t=-1$.

Therefore, $g(x) = \begin{cases} f(x)\;\;\;\; -3\leq{x}\lt-1 \\ -1 \;\;\;\;\;\; -1\leq{x}\leq{0} \end{cases}$

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CASE II

As $0\leq{x}\leq{2}$,

There will always lie a $t=0$ in $0\leq{t}\lt{2}$, such that $f(t)$ achieves its maximum as $f(t)=0$ at $t=0$.

As $2\lt{x}\leq{3}$

$f(t)$ is strictly increasing, therefore, its maximum value corresponds to the end point of the set, i.e. $x$ achieving maximum $f(x)$.

Therefore, $g(x) = \begin{cases}\; 0\;\;\;\;\;\;\; 0\leq{x}\leq2 \\ -1 \;\;\;\;\;\; 2\le{x}\leq{3} \end{cases}$

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CONCLUSION

$g(x) =\begin{cases} x^2 + 2x, & \text{$-3 \le x < -1$} \\ -1, & \text{$-1 \le x < 0$}\\ \tag{1} 0, & \text{$0\le x \le 2$}\\ x^2 - 2x, & \text{$2 < x \le 3$}\\ \end{cases}$

Answer 2

$g(x) = \begin{cases}\min\{f(t):-3 \le t \le x,& \text{$-3 \le x<0$}\} \\ \max\{f(t): 0 \le t \le x,& \text{$0\le x \le 3$}\} \end{cases}$ means, in Layman terms,

$g(x)$ is the minimum value the function $y = f(x)$ takes in the interval $[-3, x]$, when $x$ is in $[-3, 0]$;

and maximum value the function $y = f(x)$ takes in the interval $[0, x]$, when $x$ is in $[0, 3].$