For $-2 \leq x \leq 0$, we have $f(x) = \max(4-x^2, 1+x^2)$. Since $x^2 \geq 0$, we have $4-x^2 \leq 4$ and $1+x^2 \geq 1$. Therefore, $f(x) = 4-x^2$ for $-2 \leq x \leq 0$.
For $0 < x \leq 2$, we have $f(x) = \min(4-x^2, 1+x^2)$. Since $x^2 \geq 0$, we have
Show more…