Answer :
To solve this problem, we need to determine the height of the shipping box by using the given functions for volume, length, and width. Let's break down the steps:
1. Understand the Given Functions:
- The volume of the box is given by the function [tex]\( f(x) = 2x^3 + 3x^2 - 11x - 6 \)[/tex].
- The length of the box is [tex]\( g(x) = x + 3 \)[/tex].
- The width of the box is [tex]\( h(x) = x - 2 \)[/tex].
2. Calculate the Height Function:
- The formula for the volume of a rectangular prism is:
[tex]\[
\text{Volume} = \text{Length} \times \text{Width} \times \text{Height}
\][/tex]
- Therefore, to find the height, we need to divide the volume function by the product of the length and width functions:
[tex]\[
\text{Height} = \frac{f(x)}{g(x) \times h(x)}
\][/tex]
3. Perform the Division:
- First, calculate [tex]\( g(x) \times h(x) \)[/tex]:
[tex]\[
g(x) \times h(x) = (x + 3)(x - 2) = x^2 + x - 6
\][/tex]
- Now, divide the volume function [tex]\( f(x) \)[/tex] by this product:
[tex]\[
\text{Height function} = \frac{2x^3 + 3x^2 - 11x - 6}{x^2 + x - 6}
\][/tex]
4. Simplify the Height Function:
- After performing the division and simplifying, the result is:
[tex]\[
\text{Height function} = 2x + 1
\][/tex]
5. Determine the Domain:
- The domain of the height function is determined by ensuring that the denominator in the division is not zero. Since we used [tex]\( g(x) \times h(x) \)[/tex] as the denominator, we need to ensure:
[tex]\[
x^2 + x - 6 \neq 0
\][/tex]
- The expression [tex]\( x^2 + x - 6 \)[/tex] factors to [tex]\( (x - 2)(x + 3) \)[/tex], hence it is zero when [tex]\( x = 2 \)[/tex] or [tex]\( x = -3 \)[/tex].
- Since these values would make the denominator zero, they are excluded from the domain. Therefore, the domain excludes [tex]\( x = 2 \)[/tex] and [tex]\( x = -3 \)[/tex], resulting in:
[tex]\[
\text{Domain: } (-\infty, -3) \cup (-3, 2) \cup (2, \infty)
\][/tex]
Given the explanation, the height function is [tex]\( 2x + 1 \)[/tex] with a domain of [tex]\( x > 2 \)[/tex] if we assume [tex]\( x \)[/tex] only takes positive values greater than the critical value affecting division, aligning with the practical situation in many shipping contexts.
1. Understand the Given Functions:
- The volume of the box is given by the function [tex]\( f(x) = 2x^3 + 3x^2 - 11x - 6 \)[/tex].
- The length of the box is [tex]\( g(x) = x + 3 \)[/tex].
- The width of the box is [tex]\( h(x) = x - 2 \)[/tex].
2. Calculate the Height Function:
- The formula for the volume of a rectangular prism is:
[tex]\[
\text{Volume} = \text{Length} \times \text{Width} \times \text{Height}
\][/tex]
- Therefore, to find the height, we need to divide the volume function by the product of the length and width functions:
[tex]\[
\text{Height} = \frac{f(x)}{g(x) \times h(x)}
\][/tex]
3. Perform the Division:
- First, calculate [tex]\( g(x) \times h(x) \)[/tex]:
[tex]\[
g(x) \times h(x) = (x + 3)(x - 2) = x^2 + x - 6
\][/tex]
- Now, divide the volume function [tex]\( f(x) \)[/tex] by this product:
[tex]\[
\text{Height function} = \frac{2x^3 + 3x^2 - 11x - 6}{x^2 + x - 6}
\][/tex]
4. Simplify the Height Function:
- After performing the division and simplifying, the result is:
[tex]\[
\text{Height function} = 2x + 1
\][/tex]
5. Determine the Domain:
- The domain of the height function is determined by ensuring that the denominator in the division is not zero. Since we used [tex]\( g(x) \times h(x) \)[/tex] as the denominator, we need to ensure:
[tex]\[
x^2 + x - 6 \neq 0
\][/tex]
- The expression [tex]\( x^2 + x - 6 \)[/tex] factors to [tex]\( (x - 2)(x + 3) \)[/tex], hence it is zero when [tex]\( x = 2 \)[/tex] or [tex]\( x = -3 \)[/tex].
- Since these values would make the denominator zero, they are excluded from the domain. Therefore, the domain excludes [tex]\( x = 2 \)[/tex] and [tex]\( x = -3 \)[/tex], resulting in:
[tex]\[
\text{Domain: } (-\infty, -3) \cup (-3, 2) \cup (2, \infty)
\][/tex]
Given the explanation, the height function is [tex]\( 2x + 1 \)[/tex] with a domain of [tex]\( x > 2 \)[/tex] if we assume [tex]\( x \)[/tex] only takes positive values greater than the critical value affecting division, aligning with the practical situation in many shipping contexts.
