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
asked 2020-10-21

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.

Expert Answers (1)

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 \(\displaystyle{1}{<}{x}\le{2}\)
\(\displaystyle-{1}\succ{x}\ge-{2}\)
\(\displaystyle-{2}\le-{x}{<}-{1}\)
Adding 2 throughout
\(\displaystyle{0}\le-{x}+{2}{<}{1}\)
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 \(\displaystyle{\left[-{1},{1}\right]}\)

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