Let $f(x)$ be a real value function defined by $f(x)=x^2-2|x|$

Question

Let $f(x)$ be a real value function defined by $f(x)=x^2-2|x|$. If

$g(x)$$ = \begin{cases} { \mbox{minimum}: f(t): -2 \le t \le x, x \in [-2,0) \\[2ex] \mbox{maximum} : f(t) : x \le t \le 2, x \in [0,2] } \end{cases}$

I'm not sure how I should plot this. The answer of the plot shows $f(x)$ for $-2$ to $-1$ and $1$ to $2$. It is equal to $-1$ between $-1$ and $0$. And it is equal to $0$ from $0$ to $1$.

Answer 1

Steps:

• Sketch the graph of $f$ on $[-2,0]$ and $[0,2]$;
• Looking at what you sketch above, figure out the expression for $g$ by its definition; If you are confused, try some examples: what are $g(-2), g(-1.5),g(-1),g(0),g(1),g(1.5),g(2)$?
• Sketch the graph for $g$.

Here is what the graph of $f$ looks like:

Now, exercise:

Show that $$ g(x)=\begin{cases} f(x),& x\in[-2,-1]\\ -1,&x\in (-1,0)\\ 0,& x\in[0,2]. \end{cases} $$