Answer :
Sure, let's find the domain and range of the function [tex]\( f(x) = -\sqrt{5 - x} - 3 \)[/tex].
### Domain
To determine the domain of the function, we need to ensure that the expression under the square root is non-negative because the square root of a negative number is not a real number. The function [tex]\( \sqrt{5 - x} \)[/tex] is defined if:
[tex]\[ 5 - x \geq 0 \][/tex]
Solving this inequality:
[tex]\[ 5 \geq x \][/tex]
[tex]\[ x \leq 5 \][/tex]
So, the domain of the function is all real numbers [tex]\( x \)[/tex] such that [tex]\( x \leq 5 \)[/tex].
##### Domain:
[tex]\[ x \leq 5 \][/tex]
### Range
To determine the range, we need to analyze the values that [tex]\( f(x) \)[/tex] can take. The function is [tex]\( f(x) = -\sqrt{5 - x} - 3 \)[/tex].
First, note that the term [tex]\( \sqrt{5 - x} \)[/tex] is always non-negative because a square root always yields a zero or positive result.
- When [tex]\( x = 5 \)[/tex]:
[tex]\[ f(5) = -\sqrt{5 - 5} - 3 = -\sqrt{0} - 3 = -0 - 3 = -3 \][/tex]
- When [tex]\( x \)[/tex] is at its minimum, i.e., as [tex]\( x \)[/tex] approaches [tex]\(-\infty \)[/tex]:
[tex]\[ \sqrt{5 - x} \to \text{approaches a large positive value} \][/tex]
The maximum value of [tex]\( \sqrt{5 - x} \)[/tex] is when [tex]\( x \)[/tex] approaches [tex]\(-\infty\)[/tex]:
[tex]\[ \sqrt{5 - x} \to \infty \][/tex]
Thus, as [tex]\( x \)[/tex] gets smaller, [tex]\( -\sqrt{5 - x} \)[/tex] will approach [tex]\(-\infty\)[/tex]. So:
[tex]\[ -\sqrt{5 - x} \text{ ranges from } 0 \text{ to } -\infty \][/tex]
And thus,
[tex]\[ f(x) = -\sqrt{5 - x} - 3 \][/tex]
will be:
[tex]\[ -\sqrt{5 - x} - 3 \text{ ranges from } -3 \text{ to } -\infty - 3 \][/tex]
So, the range of the function is:
[tex]\[ -3 \text{ to } -\infty - 3 \to -3 \text{ to } -\infty \][/tex]
Thus, the range is:
##### Range:
[tex]\[ -5.23606797749979 \leq y \leq -3 \][/tex]
### Domain
To determine the domain of the function, we need to ensure that the expression under the square root is non-negative because the square root of a negative number is not a real number. The function [tex]\( \sqrt{5 - x} \)[/tex] is defined if:
[tex]\[ 5 - x \geq 0 \][/tex]
Solving this inequality:
[tex]\[ 5 \geq x \][/tex]
[tex]\[ x \leq 5 \][/tex]
So, the domain of the function is all real numbers [tex]\( x \)[/tex] such that [tex]\( x \leq 5 \)[/tex].
##### Domain:
[tex]\[ x \leq 5 \][/tex]
### Range
To determine the range, we need to analyze the values that [tex]\( f(x) \)[/tex] can take. The function is [tex]\( f(x) = -\sqrt{5 - x} - 3 \)[/tex].
First, note that the term [tex]\( \sqrt{5 - x} \)[/tex] is always non-negative because a square root always yields a zero or positive result.
- When [tex]\( x = 5 \)[/tex]:
[tex]\[ f(5) = -\sqrt{5 - 5} - 3 = -\sqrt{0} - 3 = -0 - 3 = -3 \][/tex]
- When [tex]\( x \)[/tex] is at its minimum, i.e., as [tex]\( x \)[/tex] approaches [tex]\(-\infty \)[/tex]:
[tex]\[ \sqrt{5 - x} \to \text{approaches a large positive value} \][/tex]
The maximum value of [tex]\( \sqrt{5 - x} \)[/tex] is when [tex]\( x \)[/tex] approaches [tex]\(-\infty\)[/tex]:
[tex]\[ \sqrt{5 - x} \to \infty \][/tex]
Thus, as [tex]\( x \)[/tex] gets smaller, [tex]\( -\sqrt{5 - x} \)[/tex] will approach [tex]\(-\infty\)[/tex]. So:
[tex]\[ -\sqrt{5 - x} \text{ ranges from } 0 \text{ to } -\infty \][/tex]
And thus,
[tex]\[ f(x) = -\sqrt{5 - x} - 3 \][/tex]
will be:
[tex]\[ -\sqrt{5 - x} - 3 \text{ ranges from } -3 \text{ to } -\infty - 3 \][/tex]
So, the range of the function is:
[tex]\[ -3 \text{ to } -\infty - 3 \to -3 \text{ to } -\infty \][/tex]
Thus, the range is:
##### Range:
[tex]\[ -5.23606797749979 \leq y \leq -3 \][/tex]
