Step-by-step explanation:
Initially, let the side of the cubical box is [tex]a[/tex] inches,
So, the volume of the box, [tex]x = a \times a \times a= a ^3[/tex].
Given that [tex]g(x)=\sqrt[3]{x} =\sqrt[3]{a^3}[/tex]
[tex]\Rightarrow g(x)=a \; \cdots (i)[/tex]
On doubling the volume of the box, the new volume is [tex]2x[/tex].
So, [tex]g(x)=\sqrt[3]{2x} =\sqrt[3]{2a^3}[/tex]
[tex]\Rightarrow g(x)=\sqrt[3]{2} a \; \cdots (i)[/tex]
As [tex]\sqrt[3]{2} >1[/tex], so the graph of [tex]g(x)=\sqrt[3]{2} a[/tex] will be above of the [tex]g(x)=a[/tex] as shown in the graph.