Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Join our Q&A platform and get accurate answers to all your questions from professionals across multiple disciplines. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
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].
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.
