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Sagot :
To find the domain of the function [tex]\( f(x) = (x-1)^{\frac{1}{2}} \)[/tex], let's analyze the expression inside the square root. This function involves a square root, which is defined only for non-negative values. Therefore, the expression inside the square root, [tex]\( x-1 \)[/tex], must be greater than or equal to zero.
Here are the steps:
1. Set up the inequality: For the function [tex]\( f(x) \)[/tex] to be real-valued and defined, the argument of the square root must be non-negative. Hence,
[tex]\[ x - 1 \geq 0. \][/tex]
2. Solve the inequality: Solve for [tex]\( x \)[/tex] in the inequality.
[tex]\[ x - 1 \geq 0 \implies x \geq 1. \][/tex]
3. Express the solution in interval notation: The inequality [tex]\( x \geq 1 \)[/tex] means that [tex]\( x \)[/tex] can be any number starting from 1 up to infinity.
Therefore, the domain of the function [tex]\( f(x) = (x-1)^{\frac{1}{2}} \)[/tex] is:
[tex]\[ [1, \infty) \][/tex]
Here are the steps:
1. Set up the inequality: For the function [tex]\( f(x) \)[/tex] to be real-valued and defined, the argument of the square root must be non-negative. Hence,
[tex]\[ x - 1 \geq 0. \][/tex]
2. Solve the inequality: Solve for [tex]\( x \)[/tex] in the inequality.
[tex]\[ x - 1 \geq 0 \implies x \geq 1. \][/tex]
3. Express the solution in interval notation: The inequality [tex]\( x \geq 1 \)[/tex] means that [tex]\( x \)[/tex] can be any number starting from 1 up to infinity.
Therefore, the domain of the function [tex]\( f(x) = (x-1)^{\frac{1}{2}} \)[/tex] is:
[tex]\[ [1, \infty) \][/tex]
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