Question # 1)y={(sqrt(-x), -4 <= x <= 0),(sqrt x, 0 < x <= 4):}2)y={(-x-2, -2 <= x <=-1),(x, -1<x<=1),(-x+2, 1 < x <=2):}Piecewise-Defined Functions. Find the (a) domain and (b) range.

Piecewise-Defined Functions
ANSWERED 1)$$\displaystyle{y}={\left\lbrace\begin{array}{cc} \sqrt{{-{x}}}&-{4}\le{x}\le{0}\\\sqrt{{x}}&{0}{<}{x}\le{4}\end{array}\right.}$$
2)$$\displaystyle{y}={\left\lbrace\begin{array}{cc} -{x}-{2}&-{2}\le{x}\le-{1}\\{x}&-{1}{<}{x}\le{1}\\-{x}+{2}&{1}{<}{x}\le{2}\end{array}\right.}$$
Piecewise-Defined Functions. Find the (a) domain and (b) range. 2020-10-22

Step 1
1)$$\displaystyle{y}={\left\lbrace\begin{array}{cc} \sqrt{{-{x}}}&-{4}\le{x}\le{0}\\\sqrt{{x}}&{0}{<}{x}\le{4}\end{array}\right.}$$
For $$\displaystyle{x}\ge{0}\sqrt{{x}}$$ is real, for $$\displaystyle{x}\le{0},\sqrt{{-{x}}}$$ is real.
y is real for $$\displaystyle\forall{X}\in{\left[-{4},{4}\right]}$$.
Hence the domain of the functions is $$\displaystyle{\left[-{4},{4}\right]}$$
When $$\displaystyle{x}\in{\left({0},{4}\right]},\sqrt{{x}}\in{\left({0},{2}\right]}$$
When $$\displaystyle{x}\in{\left[-{4},{0}\right]},-{x}\in{\left[{0},{4}\right]}$$, hence $$\displaystyle\sqrt{{-{x}}}\in{\left[{0},{2}\right]}$$
The set of all values y takes is $$\displaystyle{\left[{0},{2}\right]}$$
Hence the range is $$\displaystyle{\left[{0},{2}\right]}$$
Step 2
2)$$\displaystyle{y}={\left\lbrace\begin{array}{cc} -{x}-{2}&-{2}\le{x}\le-{1}\\{x}&-{1}{<}{x}\le{1}\\-{x}+{2}&{1}{<}{x}\le{2}\end{array}\right.}$$
The given functions is defined for $$\displaystyle{x}\in{\left[-{2},{2}\right]}$$
Hence the domain of the function is $$\displaystyle{\left[-{2},{2}\right]}$$
Range
i) When $$\displaystyle-{2}\le{x}\le-{1}$$
Multiplying by -1 throughput the inequality,
$$\displaystyle{2}\ge-{x}\ge-{1}$$
$$\displaystyle{1}\le-{x}\le{2}$$
Subtracting by 2 through out
$$\displaystyle-{1}\le-{x}-{2}\le{0}$$
So, $$\displaystyle{y}={\left(-{x}-{2}\right)}\in{\left[-{1},{0}\right]}$$
ii) When $$\displaystyle{x}\in{\left[-{1},{1}\right]},{y}={x}$$, hence $$\displaystyle{y}\in{\left[-{1},{1}\right]}$$
iii) When $$\displaystyle{1}{<}{x}\le{2}$$
$$\displaystyle-{1}\succ{x}\ge-{2}$$
$$\displaystyle-{2}\le-{x}{<}-{1}$$
$$\displaystyle{0}\le-{x}+{2}{<}{1}$$
Hence when $$\displaystyle{1}{<}{x}\le{2},{y}\in{\left[{0},{1}\right)}$$
So, the set of all values that y takes is $$\displaystyle{\left[-{1},{1}\right]}$$
Hence the range is $$\displaystyle{\left[-{1},{1}\right]}$$