Question

which function has a graph with domain $x \geq -1$ and range $y \geq -2$?

choose the correct answer below.

\\(\bigcirc\\) a. \\( f(x) = \sqrt{x + 1} - 2 \\)
\\(\bigcirc\\) b. \\( f(x) = \sqrt3{x + 1} - 2 \\)
\\(\bigcirc\\) c. \\( f(x) = \sqrt3{x - 1} + 2 \\)
\\(\bigcirc\\) d. \\( f(x) = \sqrt{x - 1} + 2 \\)

Explanation:

Step1: Analyze Domain of Square Root and Cube Root

For a square root function $\sqrt{u}$, the domain requires $u\geq0$. For a cube root function $\sqrt[3]{u}$, the domain is all real numbers (since cube of negative is negative, cube of positive is positive, and cube of 0 is 0).

Step2: Analyze Option A

Function $f(x)=\sqrt{x + 1}-2$. For the square root, $x + 1\geq0\implies x\geq - 1$ (domain). The square root $\sqrt{x + 1}\geq0$, so $\sqrt{x + 1}-2\geq - 2$ (range $y\geq - 2$).

Step3: Analyze Option B

Function $f(x)=\sqrt[3]{x + 1}-2$. Cube root has domain all real numbers (not $x\geq - 1$), so eliminate B.

Step4: Analyze Option C

Function $f(x)=\sqrt[3]{x - 1}+2$. Cube root domain all real numbers, and domain not $x\geq - 1$, eliminate C.

Step5: Analyze Option D

Function $f(x)=\sqrt{x - 1}+2$. For square root, $x - 1\geq0\implies x\geq1$ (domain not $x\geq - 1$), eliminate D.

Answer:

A. $f(x)=\sqrt{x + 1}-2$