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.
Best Answer
$g(x) = \begin{cases}\min\{f(t):-3 \le t \le x,& \text{$-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 $[0, 3].$