Question

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

Explanation:

Step1: Recall domain of square root

For \( y = \sqrt{x} \), the expression under the square root (radicand) must be non - negative. So we set up the inequality \( x\geq0 \).

Step2: Analyze the inequality

The inequality \( x\geq0 \) means that \( x \) can be equal to 0 or any positive real number. In interval notation, this is \( 0\leq x<\infty \). Now we check the options:

• The first option \( -\infty < x < \infty \) is the domain of linear functions or polynomials of even degree with no restrictions, but for square root, we can't have negative numbers.
• The second option \( 0 < x < \infty \) excludes 0, but \( \sqrt{0}=0 \) is a valid output, so 0 should be included.
• The third option \( 0\leq x < \infty \) includes 0 and all positive real numbers, which satisfies the condition for the square root function.
• The fourth option \( 1\leq x < \infty \) is incorrect because \( x = 0 \) is a valid input (\( \sqrt{0}=0 \)).

Answer:

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