Question

he models the volume, ( v(x) ), in cubic feet, as a function of the length ( x ), in feet. ( v(x) = 2x(x - 1)^2 ) he created the diagram and graph to model the box and its volume. the zeros of function ( v ) occur at . to make the box, samson should use a value for ( x ).

Explanation:

Step1: Find zeros of \( v(x) \)

To find the zeros of \( v(x) = 2x(x - 1)^2 \), set \( v(x) = 0 \).
Using the zero - product property: if \( ab = 0 \), then \( a = 0 \) or \( b = 0 \).
For \( 2x(x - 1)^2=0 \), we have two cases:

• Case 1: \( 2x = 0 \), which gives \( x = 0 \).
• Case 2: \( (x - 1)^2=0 \), taking the square root of both sides, \( x - 1=0 \), so \( x = 1 \).

Step2: Determine valid \( x \) for the box

Since \( x \) represents the length of a box, it must be a positive quantity. Also, from the diagram (with dimensions 2 ft and 4 ft related to the box - making process), we know that \( x>0 \) and \( x
eq1 \) (because if \( x = 1 \), the volume \( v(1)=2\times1\times(1 - 1)^2 = 0 \), which does not make sense for a box with positive volume). We need to choose a value of \( x \) such that \( x>0 \) and \( x
eq1 \). A valid value could be \( x = 0.5 \) (or any other value in the interval \( (0,1)\cup(1,\infty) \), for example, \( x = 2 \) is also valid as long as it fits the physical constraints of the box - making process). But for the zeros, the zeros of the function \( v(x) \) occur at \( x = 0 \) and \( x = 1 \). For the value of \( x \) to make the box, we need \( x>0 \) and \( x
eq1 \), so we can choose a value like \( x = 0.5 \) (or \( x = 2 \), etc.).

Answer:

The zeros of the function \( v(x) \) occur at \( x = 0 \) and \( x = 1 \). To make the box, Samson should use a value of \( x \) such that \( x>0 \) and \( x
eq1 \) (for example, \( x = 0.5 \) or \( x = 2 \), etc.). If we are only asked for the zeros: The zeros are \( x = 0 \) and \( x = 1 \). If we are asked for a valid \( x \) to make the box, a valid value is \( x = 0.5 \) (or other positive values except \( x = 1 \)).