Question

what is the domain of the function $y = \sqrt{x}$?
$0 < x < \infty$
$1 \leq x < \infty$
$-\infty < x < \infty$
$0 \leq x < \infty$

Explanation:

Step1: Recall domain of square root function

The square root function \( y = \sqrt{x} \) is defined when the expression inside the square root (the radicand) is non - negative. So we need \( x\geq0 \).

Step2: Analyze the options

• For the option \( 0 < x<\infty \), it excludes \( x = 0 \), but \( \sqrt{0}=0 \) is a valid output, so this is incorrect.
• For the option \( 1\leq x<\infty \), it excludes values between \( 0 \) and \( 1 \) (excluding \( 1 \)) like \( x = 0.25 \) where \( \sqrt{0.25}=0.5 \) is valid, so this is incorrect.
• For the option \( -\infty < x<\infty \), it includes negative numbers, but the square root of a negative number is not a real number (in the context of real - valued functions, which is the usual context for this type of problem), so this is incorrect.
• For the option \( 0\leq x<\infty \), it includes all non - negative real numbers, which is the set of values for which \( y=\sqrt{x} \) is defined.

Answer:

\( 0\leq x<\infty \) (the option with the inequality \( 0\leq x<\infty \))