Question

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

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 to find all real numbers \( x \) such that \( x\geq0 \).

Step2: Analyze the options

• The first option \( -\infty < x < \infty \) is the domain of functions like linear functions \( y = mx + b \), but for square root, we can't have negative \( x \), so this is wrong.
• The second option \( 0 < x < \infty \) excludes \( x = 0 \). But when \( x = 0 \), \( y=\sqrt{0}=0 \) is defined, so this is wrong.
• The third option \( 0\leq x < \infty \) includes all non - negative real numbers, which is correct because when \( x = 0 \), the function is defined (\( y = 0 \)) and for all \( x>0 \), the square root is defined in the real number system.
• The fourth option \( 1\leq x < \infty \) excludes values between \( 0 \) and \( 1 \). For example, when \( x=\frac{1}{4} \), \( y = \sqrt{\frac{1}{4}}=\frac{1}{2} \) is defined, so this is wrong.

Answer:

\( 0\leq x < \infty \) (the third option)