Answer :
To solve the problem of computing the square roots involving negative [tex]\( x \)[/tex] and different shifts, we need to evaluate [tex]\(\sqrt{-x + \text{shift}}\)[/tex] for each given shift. We will show our work step-by-step for each shift value.
### Negative Shifts
#### Shift: -5
[tex]\[ \sqrt{-x - 5} \][/tex]
For [tex]\(\sqrt{-x - 5}\)[/tex] to be a real number, the expression inside the square root must be non-negative:
[tex]\[-x - 5 \geq 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ -x \geq 5 \implies x \leq -5 \][/tex]
We can provide a specific example with [tex]\( x = -5 \)[/tex]:
[tex]\[ \sqrt{-(-5) - 5} = \sqrt{5 - 5} = \sqrt{0} = 0 \][/tex]
Thus, [tex]\(\sqrt{-x - 5}\)[/tex] yields a real number (0) when [tex]\( x = -5 \)[/tex].
#### Shift: -4
[tex]\[ \sqrt{-x - 4} \][/tex]
For [tex]\(\sqrt{-x - 4}\)[/tex] to be real:
[tex]\[-x - 4 \geq 0 \][/tex]
[tex]\[ -x \geq 4 \implies x \leq -4 \][/tex]
Taking [tex]\( x = -4 \)[/tex]:
[tex]\[ \sqrt{-(-4) - 4} = \sqrt{4 - 4} = \sqrt{0} = 0 \][/tex]
Thus, [tex]\(\sqrt{-x - 4}\)[/tex] yields a real number (0) when [tex]\( x = -4 \)[/tex].
#### Shift: -3
[tex]\[ \sqrt{-x - 3} \][/tex]
For [tex]\(\sqrt{-x - 3}\)[/tex] to be real:
[tex]\[-x - 3 \geq 0 \][/tex]
[tex]\[ -x \geq 3 \implies x \leq -3 \][/tex]
Taking [tex]\( x = -3 \)[/tex]:
[tex]\[ \sqrt{-(-3) - 3} = \sqrt{3 - 3} = \sqrt{0} = 0 \][/tex]
Thus, [tex]\(\sqrt{-x - 3}\)[/tex] yields a real number (0) when [tex]\( x = -3 \)[/tex].
#### Shift: -2
[tex]\[ \sqrt{-x - 2} \][/tex]
For [tex]\(\sqrt{-x - 2}\)[/tex] to be real:
[tex]\[-x - 2 \geq 0 \][/tex]
[tex]\[ -x \geq 2 \implies x \leq -2 \][/tex]
Taking [tex]\( x = -2 \)[/tex]:
[tex]\[ \sqrt{-(-2) - 2} = \sqrt{2 - 2} = \sqrt{0} = 0 \][/tex]
Thus, [tex]\(\sqrt{-x - 2}\)[/tex] yields a real number (0) when [tex]\( x = -2 \)[/tex].
### Positive Shifts
#### Shift: 1
[tex]\[ \sqrt{-x + 1} \][/tex]
For [tex]\(\sqrt{-x + 1}\)[/tex] to be real:
[tex]\[-x + 1 \geq 0 \][/tex]
[tex]\[ -x \geq -1 \implies x \leq 1 \][/tex]
Taking [tex]\( x = 0 \)[/tex]:
[tex]\[ \sqrt{-0 + 1} = \sqrt{1} = 1 \][/tex]
Thus, [tex]\(\sqrt{-x + 1}\)[/tex] yields a real number (1) when [tex]\( x = 0 \)[/tex].
#### Shift: 2
[tex]\[ \sqrt{-x + 2} \][/tex]
For [tex]\(\sqrt{-x + 2}\)[/tex] to be real:
[tex]\[-x + 2 \geq 0 \][/tex]
[tex]\[ -x \geq -2 \implies x \leq 2 \][/tex]
Taking [tex]\( x = 0 \)[/tex]:
[tex]\[ \sqrt{-0 + 2} = \sqrt{2} \][/tex]
Thus, [tex]\(\sqrt{-x + 2}\)[/tex] yields a real number ([tex]\(\sqrt{2}\)[/tex]) when [tex]\( x = 0 \)[/tex].
#### Shift: 3
[tex]\[ \sqrt{-x + 3} \][/tex]
For [tex]\(\sqrt{-x + 3}\)[/tex] to be real:
[tex]\[-x + 3 \geq 0 \][/tex]
[tex]\[ -x \geq -3 \implies x \leq 3 \][/tex]
Taking [tex]\( x = 0 \)[/tex]:
[tex]\[ \sqrt{-0 + 3} = \sqrt{3} \][/tex]
Thus, [tex]\(\sqrt{-x + 3}\)[/tex] yields a real number ([tex]\(\sqrt{3}\)[/tex]) when [tex]\( x = 0 \)[/tex].
#### Shift: 4
[tex]\[ \sqrt{-x + 4} \][/tex]
For [tex]\(\sqrt{-x + 4}\)[/tex] to be real:
[tex]\[-x + 4 \geq 0 \][/tex]
[tex]\[ -x \geq -4 \implies x \leq 4 \][/tex]
Taking [tex]\( x = 0 \)[/tex]:
[tex]\[ \sqrt{-0 + 4} = \sqrt{4} = 2 \][/tex]
Thus, [tex]\(\sqrt{-x + 4}\)[/tex] yields a real number (2) when [tex]\( x = 0 \)[/tex].
#### Shift: 5
[tex]\[ \sqrt{-x + 5} \][/tex]
For [tex]\(\sqrt{-x + 5}\)[/tex] to be real:
[tex]\[-x + 5 \geq 0 \][/tex]
[tex]\[ -x \geq -5 \implies x \leq 5 \][/tex]
Taking [tex]\( x = 0 \)[/tex]:
[tex]\[ \sqrt{-0 + 5} = \sqrt{5} \][/tex]
Thus, [tex]\(\sqrt{-x + 5}\)[/tex] yields a real number ([tex]\(\sqrt{5}\)[/tex]) when [tex]\( x = 0 \)[/tex].
### Negative Shifts
#### Shift: -5
[tex]\[ \sqrt{-x - 5} \][/tex]
For [tex]\(\sqrt{-x - 5}\)[/tex] to be a real number, the expression inside the square root must be non-negative:
[tex]\[-x - 5 \geq 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ -x \geq 5 \implies x \leq -5 \][/tex]
We can provide a specific example with [tex]\( x = -5 \)[/tex]:
[tex]\[ \sqrt{-(-5) - 5} = \sqrt{5 - 5} = \sqrt{0} = 0 \][/tex]
Thus, [tex]\(\sqrt{-x - 5}\)[/tex] yields a real number (0) when [tex]\( x = -5 \)[/tex].
#### Shift: -4
[tex]\[ \sqrt{-x - 4} \][/tex]
For [tex]\(\sqrt{-x - 4}\)[/tex] to be real:
[tex]\[-x - 4 \geq 0 \][/tex]
[tex]\[ -x \geq 4 \implies x \leq -4 \][/tex]
Taking [tex]\( x = -4 \)[/tex]:
[tex]\[ \sqrt{-(-4) - 4} = \sqrt{4 - 4} = \sqrt{0} = 0 \][/tex]
Thus, [tex]\(\sqrt{-x - 4}\)[/tex] yields a real number (0) when [tex]\( x = -4 \)[/tex].
#### Shift: -3
[tex]\[ \sqrt{-x - 3} \][/tex]
For [tex]\(\sqrt{-x - 3}\)[/tex] to be real:
[tex]\[-x - 3 \geq 0 \][/tex]
[tex]\[ -x \geq 3 \implies x \leq -3 \][/tex]
Taking [tex]\( x = -3 \)[/tex]:
[tex]\[ \sqrt{-(-3) - 3} = \sqrt{3 - 3} = \sqrt{0} = 0 \][/tex]
Thus, [tex]\(\sqrt{-x - 3}\)[/tex] yields a real number (0) when [tex]\( x = -3 \)[/tex].
#### Shift: -2
[tex]\[ \sqrt{-x - 2} \][/tex]
For [tex]\(\sqrt{-x - 2}\)[/tex] to be real:
[tex]\[-x - 2 \geq 0 \][/tex]
[tex]\[ -x \geq 2 \implies x \leq -2 \][/tex]
Taking [tex]\( x = -2 \)[/tex]:
[tex]\[ \sqrt{-(-2) - 2} = \sqrt{2 - 2} = \sqrt{0} = 0 \][/tex]
Thus, [tex]\(\sqrt{-x - 2}\)[/tex] yields a real number (0) when [tex]\( x = -2 \)[/tex].
### Positive Shifts
#### Shift: 1
[tex]\[ \sqrt{-x + 1} \][/tex]
For [tex]\(\sqrt{-x + 1}\)[/tex] to be real:
[tex]\[-x + 1 \geq 0 \][/tex]
[tex]\[ -x \geq -1 \implies x \leq 1 \][/tex]
Taking [tex]\( x = 0 \)[/tex]:
[tex]\[ \sqrt{-0 + 1} = \sqrt{1} = 1 \][/tex]
Thus, [tex]\(\sqrt{-x + 1}\)[/tex] yields a real number (1) when [tex]\( x = 0 \)[/tex].
#### Shift: 2
[tex]\[ \sqrt{-x + 2} \][/tex]
For [tex]\(\sqrt{-x + 2}\)[/tex] to be real:
[tex]\[-x + 2 \geq 0 \][/tex]
[tex]\[ -x \geq -2 \implies x \leq 2 \][/tex]
Taking [tex]\( x = 0 \)[/tex]:
[tex]\[ \sqrt{-0 + 2} = \sqrt{2} \][/tex]
Thus, [tex]\(\sqrt{-x + 2}\)[/tex] yields a real number ([tex]\(\sqrt{2}\)[/tex]) when [tex]\( x = 0 \)[/tex].
#### Shift: 3
[tex]\[ \sqrt{-x + 3} \][/tex]
For [tex]\(\sqrt{-x + 3}\)[/tex] to be real:
[tex]\[-x + 3 \geq 0 \][/tex]
[tex]\[ -x \geq -3 \implies x \leq 3 \][/tex]
Taking [tex]\( x = 0 \)[/tex]:
[tex]\[ \sqrt{-0 + 3} = \sqrt{3} \][/tex]
Thus, [tex]\(\sqrt{-x + 3}\)[/tex] yields a real number ([tex]\(\sqrt{3}\)[/tex]) when [tex]\( x = 0 \)[/tex].
#### Shift: 4
[tex]\[ \sqrt{-x + 4} \][/tex]
For [tex]\(\sqrt{-x + 4}\)[/tex] to be real:
[tex]\[-x + 4 \geq 0 \][/tex]
[tex]\[ -x \geq -4 \implies x \leq 4 \][/tex]
Taking [tex]\( x = 0 \)[/tex]:
[tex]\[ \sqrt{-0 + 4} = \sqrt{4} = 2 \][/tex]
Thus, [tex]\(\sqrt{-x + 4}\)[/tex] yields a real number (2) when [tex]\( x = 0 \)[/tex].
#### Shift: 5
[tex]\[ \sqrt{-x + 5} \][/tex]
For [tex]\(\sqrt{-x + 5}\)[/tex] to be real:
[tex]\[-x + 5 \geq 0 \][/tex]
[tex]\[ -x \geq -5 \implies x \leq 5 \][/tex]
Taking [tex]\( x = 0 \)[/tex]:
[tex]\[ \sqrt{-0 + 5} = \sqrt{5} \][/tex]
Thus, [tex]\(\sqrt{-x + 5}\)[/tex] yields a real number ([tex]\(\sqrt{5}\)[/tex]) when [tex]\( x = 0 \)[/tex].
