Answer:
The range and domain of this function is [-2, 2]
Step-by-step explanation:
For the domain it must be fulfilled that [tex] (4-x^2)\geq 0[/tex]. Then, [tex](2 + x) (2-x)\geq 0[/tex]. This expression is true for the values of [tex]x[/tex] that make the factors simultaneously non-negative or non-positive.
Case non-negative factors
[tex]2 + x \geq 0[/tex] implies that [tex]x \geq -2[/tex], that is [tex][-2, +\infty][/tex]
[tex]2-x \geq 0[/tex] implies that [tex]x \leq 2[/tex], that is, [tex](-\infty, 2][/tex]. The intersection of the previous sets is [-2, 2].
Case non positive factors
[tex]2 + x \leq 0[/tex] implies that [tex]x \leq -2[/tex], that is [tex](-\infty, -2][/tex]
[tex]2-x \leq 0[/tex] implies that [tex]2 \leq x[/tex], that is, [tex][2,+\infty)[/tex]. The intersection of the previous sets is empty.
Then the domain of the function is [-2, 2]
The range of this function is the domain of its inverse. The inverse of the function is itself, so the range is [-2, 2]
