Let's analyze the function [tex]\( f(x) = \sqrt{1 - 2x} \)[/tex].
### Step 1: Determining the Domain
The domain of a function consists of all the possible input values (x-values) that make the function's expression valid. Since [tex]\( f(x) \)[/tex] involves a square root, the expression inside the square root must be non-negative (i.e., greater than or equal to zero).
[tex]\[ 1 - 2x \geq 0 \][/tex]
Solving this inequality for [tex]\( x \)[/tex]:
1. Subtract 1 from both sides:
[tex]\[ -2x \geq -1 \][/tex]
2. Divide both sides by -2 (and remember to reverse the inequality sign when dividing by a negative number):
[tex]\[ x \leq \frac{1}{2} \][/tex]
Therefore, the domain is:
[tex]\[ x \in (-\infty, \frac{1}{2}] \][/tex]
### Step 2: Determining the Range
The range of a function consists of all the possible output values (y-values). Since [tex]\( f(x) = \sqrt{1 - 2x} \)[/tex], we need to consider the values that the square root can take.
1. The square root function, [tex]\( \sqrt{u} \)[/tex], produces non-negative values (i.e., [tex]\( \sqrt{u} \geq 0 \)[/tex]).
2. The expression inside the square root, [tex]\( 1 - 2x \)[/tex], reaches its maximum value when [tex]\( x \)[/tex] is at the smallest value in the domain. This occurs at [tex]\( x = -\infty \)[/tex], theoretically, making the expression tend to 1.
So, when [tex]\( x = -\infty \)[/tex]:
[tex]\[ f(x) = \sqrt{1} = 1 \][/tex]
As [tex]\( x \)[/tex] increases from [tex]\( -\infty \)[/tex] up to [tex]\( \frac{1}{2} \)[/tex], the expression [tex]\( 1 - 2x \)[/tex] decreases from 1 to 0, making the function [tex]\( \sqrt{1 - 2x} \)[/tex] range from 1 down to 0.
Therefore, the range of [tex]\( f(x) \)[/tex] is:
[tex]\[ y \in [0, 1] \][/tex]
### Conclusion
The function [tex]\( f(x) = \sqrt{1 - 2x} \)[/tex] has the following domain and range:
- Domain: [tex]\( (-\infty, \frac{1}{2}] \)[/tex]
- Range: [tex]\( [0, 1] \)[/tex]
