To find the height of the shipping box, we need to relate the volume, length, and width. Given:
- Volume: \( f(x) = 2x^3 + 3x^2 - 11x - 6 \)
- Length: \( g(x) = x + 3 \)
- Width: \( h(x) = x - 2 \)
The height \( k(x) \) of the box can be expressed in terms of the volume, length, and width:
\[ k(x) = \frac{f(x)}{g(x) \cdot h(x)} \]
First, calculate \( g(x) \times h(x) \):
\[ g(x) \cdot h(x) = (x + 3)(x - 2) = x^2 + 3x - 2x - 6 = x^2 + x - 6 \]
Now, we need to divide \( f(x) \) by \( g(x) \cdot h(x) \):
The height function becomes:
\[ k(x) = \frac{f(x)}{g(x) \cdot h(x)} = \frac{2x^3 + 3x^2 - 11x - 6}{x^2 + x - 6} \]
Next, perform polynomial long division to simplify this expression.
- Divide \( 2x^3 \) by \( x^2 \) to get \( 2x \).
- Multiply \( 2x \) by \( x^2 + x - 6 \) and subtract from \( f(x) \): \[ 2x(x^2 + x - 6) = 2x^3 + 2x^2 - 12x \] Subtract: \[ (2x^3 + 3x^2 - 11x - 6) - (2x^3 + 2x^2 - 12x) = (3x^2 - 2x^2) + (-11x + 12x) - 6 = x^2 + x - 6 \]
- Now divide \( x^2 \) by \( x^2 \) to get \( 1 \). Multiply by \( x^2 + x - 6 \) and subtract: \[ 1(x^2 + x - 6) = x^2 + x - 6 \] Subtract again: \[ (x^2 + x - 6) - (x^2 + x - 6) = 0 \]
The result of the polynomial division shows that:
\[ k(x) = 2x + 1 \]
Next, consider the domain. The volume, length, and width cannot be negative, and since the length is \( g(x) = x + 3 \) and the width is \( h(x) = x - 2 \), both must be positive:
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For length \( g(x) > 0 \): \[ x + 3 > 0 \rightarrow x > -3 \]
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For width \( h(x) > 0 \): \[ x - 2 > 0 \rightarrow x > 2 \]
Combining these constraints, the domain of \( k(x) \) is:
\[ x > 2 \]
Therefore, the height function is:
\[ k(x) = 2x + 1 \]
And the domain is \( x > 2 \).
So, the correct answer is:
\((fg \cdot h)(x) = 2x + 1\) and the domain is \( x > 2\).
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