Question

the graph shows the function g(x) for a restricted domain
which is the function g(x) for a restricted domain?
\\( g(x) = \sqrt3{x - 4}; x \geq -4 \\)
\\( g(x) = \sqrt3{x} + 4; x \geq 0 \\)
\\( g(x) = \sqrt3{x + 4}; x \geq -4 \\)
\\( g(x) = \sqrt3{x} - 4; x \geq 0 \\)

Explanation:

Step1: Analyze the domain from the graph

The graph starts at \( x = -4 \) (the leftmost point is at \( x = -4 \)), so the domain is \( x \geq -4 \). This eliminates options with domain \( x \geq 0 \) (second and fourth options).

Step2: Analyze the function form

The parent cube root function is \( y=\sqrt[3]{x} \). A horizontal shift of \( h \) units is \( y = \sqrt[3]{x - h} \) (right shift) or \( y=\sqrt[3]{x + h} \) (left shift). Here, the graph seems to be a left shift? Wait, no—when \( x=-4 \), let's check the function. For the third option: \( g(x)=\sqrt[3]{x + 4} \), when \( x=-4 \), \( g(-4)=\sqrt[3]{-4 + 4}=0 \), which matches the graph's starting point (at \( x=-4 \), \( y = 0 \)? Wait, no, the graph at \( x=-4 \) has \( y \) starting, let's see the graph: at \( x=-4 \), the point is \( (-4, 0) \)? Wait, the graph starts at \( x=-4 \), and when \( x = 0 \), let's check the third option: \( g(0)=\sqrt[3]{0 + 4}=\sqrt[3]{4}\approx1.587 \), but the graph at \( x=0 \) is around \( y = 2 \)? Wait, maybe my initial thought. Wait, the first option: \( g(x)=\sqrt[3]{x - 4} \), domain \( x\geq -4 \)? No, the domain for cube root is all real numbers, but here restricted. Wait, no—wait the third option: \( g(x)=\sqrt[3]{x + 4} \), domain \( x\geq -4 \). Let's check the vertex (the starting point) is at \( x=-4 \), because inside the cube root, \( x + 4 \geq 0 \) (wait, no, cube root is defined for all real numbers, but maybe the restricted domain is \( x\geq -4 \) to make it a function? Wait, no, cube root is a function for all real numbers. But the graph starts at \( x=-4 \), so the domain is \( x\geq -4 \). Now, check the function: when \( x=-4 \), \( g(-4)=\sqrt[3]{-4 + 4}=0 \), which matches the graph's leftmost point (at \( x=-4 \), \( y = 0 \)? Wait, the graph at \( x=-4 \) looks like it's starting, maybe the y - value at \( x=-4 \) is 0, and as \( x \) increases, \( y \) increases. Let's check the third option: \( g(x)=\sqrt[3]{x + 4} \), domain \( x\geq -4 \). Let's test \( x = 0 \): \( \sqrt[3]{4}\approx1.587 \), but the graph at \( x=0 \) is around \( y = 2 \)? Wait, maybe I made a mistake. Wait, the second option: \( g(x)=\sqrt[3]{x}+4 \), domain \( x\geq 0 \). But the graph starts at \( x=-4 \), so domain can't be \( x\geq 0 \). First option: \( g(x)=\sqrt[3]{x - 4} \), domain \( x\geq -4 \). When \( x=-4 \), \( g(-4)=\sqrt[3]{-4 - 4}=\sqrt[3]{-8}=-2 \), which doesn't match. Third option: \( g(x)=\sqrt[3]{x + 4} \), domain \( x\geq -4 \). When \( x=-4 \), \( g(-4)=0 \); when \( x = 0 \), \( g(0)=\sqrt[3]{4}\approx1.587 \), but the graph at \( x=0 \) is around \( y = 2 \). Wait, maybe the graph is \( g(x)=\sqrt[3]{x + 4} \), and the domain is \( x\geq -4 \). Let's check the options again. The third option is \( g(x)=\sqrt[3]{x + 4}; x\geq -4 \). Let's confirm the domain: the graph starts at \( x=-4 \), so domain is \( x\geq -4 \). The function inside the cube root: \( x + 4 \), so when \( x\geq -4 \), \( x + 4 \geq 0 \), but cube root is defined for all real numbers. But maybe the restricted domain is \( x\geq -4 \) to make it a certain function. So the correct option is the third one.

Step3: Eliminate other options

- Option 1: \( g(x)=\sqrt[3]{x - 4}; x\geq -4 \). At \( x=-4 \), \( g(-4)=\sqrt[3]{-8}=-2 \), but graph at \( x=-4 \) is \( y=0 \) (or near 0), so eliminate.
- Option 2: Domain \( x\geq 0 \), but graph starts at \( x=-4 \), eliminate.
- Option 4: Domain \( x\geq 0 \), graph starts at \( x=-4 \), eliminate.
- Option 3: Domain \( x\geq -4 \), at \( x=-4 \), \( g(-4)=0 \), and as \( x \) increases, \( g(x) \) increases, matching the graph.

Answer:

\( g(x) = \sqrt[3]{x + 4}; x \geq -4 \) (the third option: \( g(x)=\sqrt[3]{x + 4}; x \geq -4 \))