To find the values of \( x \) that will produce a rectangle with an area greater than 18 square inches, we can start by calculating the area \( A \) of the rectangle using the given dimensions:
- Width: \( x - 1 \)
- Length: \( 3x \)
The area \( A \) is given by:
\[ A = \text{Width} \times \text{Length} = (x - 1)(3x) \]
Expanding this expression:
\[ A = 3x(x - 1) = 3x^2 - 3x \]
We want to find when this area is greater than 18 square inches:
\[ 3x^2 - 3x > 18 \]
To solve this inequality, we first rearrange it:
\[ 3x^2 - 3x - 18 > 0 \]
Next, we can simplify it by dividing everything by 3:
\[ x^2 - x - 6 > 0 \]
Now, we can factor the quadratic expression:
\[ (x - 3)(x + 2) > 0 \]
Now we will determine the critical points by setting the factors equal to zero:
- \( x - 3 = 0 \Rightarrow x = 3 \)
- \( x + 2 = 0 \Rightarrow x = -2 \)
These critical points divide the number line into intervals. We will test the intervals to see where the product is positive:
-
Interval \( (-\infty, -2) \):
- Choose \( x = -3 \): \(( -3 - 3)( -3 + 2) = (-6)(-1) > 0\) (True)
-
Interval \( (-2, 3) \):
- Choose \( x = 0 \): \((0 - 3)(0 + 2) = (-3)(2) < 0\) (False)
-
Interval \( (3, \infty) \):
- Choose \( x = 4 \): \((4 - 3)(4 + 2) = (1)(6) > 0\) (True)
So 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:
\[ x < -2 \quad \text{or} \quad x > 3 \]
Given the provided choices, the correct response would be: all values of x such that \( x < -2 \) or \( x > 3 \).