To find the values of \( x \) that will produce a rectangle with an area greater than 18 square inches, we start with the formula for the area \( A \) of a rectangle, which is given by:
\[ A = \text{length} \times \text{width} \]
For this rectangle, the width is \( x - 1 \) inches and the length is \( 3x \) inches. Thus, we have:
\[ A = (3x)(x - 1) = 3x^2 - 3x \]
We want the area to be greater than 18 square inches:
\[ 3x^2 - 3x > 18 \]
To solve this inequality, first, we can move all terms to one side:
\[ 3x^2 - 3x - 18 > 0 \]
Next, divide each term by 3 to simplify:
\[ x^2 - x - 6 > 0 \]
Now we can factor the quadratic:
\[ (x - 3)(x + 2) > 0 \]
Next, we find the critical points by setting the factors equal to zero:
\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \] \[ x + 2 = 0 \quad \Rightarrow \quad x = -2 \]
Now, we need to test the intervals determined by these critical points: \( (-\infty, -2) \), \( (-2, 3) \), and \( (3, \infty) \).
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Testing \( x < -2 \) (e.g., \( x = -3 \)): \[ (-3 - 3)(-3 + 2) = (-6)(-1) = 6 > 0 \quad \text{(True)} \]
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Testing \( -2 < x < 3 \) (e.g., \( x = 0 \)): \[ (0 - 3)(0 + 2) = (-3)(2) = -6 < 0 \quad \text{(False)} \]
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Testing \( x > 3 \) (e.g., \( x = 4 \)): \[ (4 - 3)(4 + 2) = (1)(6) = 6 > 0 \quad \text{(True)} \]
Combining these results, the solution to the inequality \( (x - 3)(x + 2) > 0 \) is:
\[ x < -2 \quad \text{or} \quad x > 3 \]
Thus, the values of \( x \) that will produce a rectangle with an area greater than 18 square inches are:
all values of \( x \) such that \( x < -2 \) or \( x > 3 \).
Therefore, the correct response is:
all values of x such that \( x < -2 \) or \( x > 3 \).