To determine the intervals on which the function [tex]\( y = \left|2x^2 + x - 6\right| \)[/tex] is increasing or decreasing, we will carry out the following steps:
1. Find the critical points of the function [tex]\( y = \left|2x^2 + x - 6\right| \)[/tex].
2. Analyze the sign of the first derivative to determine the intervals of increase and decrease.
### Step 1: Find the critical points
First, let's analyze the expression inside the absolute value:
[tex]\[ f(x) = 2x^2 + x - 6 \][/tex]
To find the critical points, we set the derivative of [tex]\( f(x) \)[/tex] equal to zero:
[tex]\[ f'(x) = \frac{d}{dx}(2x^2 + x - 6) = 4x + 1 \][/tex]
Setting [tex]\( f'(x) = 0 \)[/tex]:
[tex]\[ 4x + 1 = 0 \][/tex]
[tex]\[ x = -\frac{1}{4} \][/tex]
So, [tex]\( x = -\frac{1}{4} \)[/tex] is one critical point. However, because the expression is inside an absolute value function, we should also consider where [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ 2x^2 + x - 6 = 0 \][/tex]
Solving this quadratic equation:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = 2 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = -6 \)[/tex].
[tex]\[ x = \frac{-1 \pm \sqrt{1 + 48}}{4} \][/tex]
[tex]\[ x = \frac{-1 \pm 7}{4} \][/tex]
[tex]\[ x = 1.5 \text{ or } x = -2 \][/tex]
So, the critical points are [tex]\( x = -2 \)[/tex] and [tex]\( x = 1.5 \)[/tex] and potentially the point where the derivative is zero, [tex]\( x = -\frac{1}{4} \)[/tex].
### Step 2: Determine intervals of increase and decrease
To analyze the behavior of [tex]\( y = \left|2x^2 + x - 6\right| \)[/tex], we need to check the intervals around the critical points [tex]\( x = -2 \)[/tex], [tex]\( x = -\frac{1}{4} \)[/tex], and [tex]\( x = 1.5 \)[/tex].
- For [tex]\( x < -2 \)[/tex]:
[tex]\( f(x) \)[/tex] is positive, because the parabola opens upwards and is positive on this interval.
- [tex]\( y \)[/tex] is increasing.
- For [tex]\( -2 < x < -\frac{1}{4} \)[/tex]:
[tex]\( f(x) \)[/tex] is negative, so [tex]\( y = \left| f(x) \right| \)[/tex] will be decreasing.
- For [tex]\( -\frac{1}{4} < x < 1.5 \)[/tex]:
[tex]\( f(x) \)[/tex] changes its increasing and decreasing behavior separately, but since we're considering the absolute values, the value will be increasing as it changes its direction beyond zero.
- For [tex]\( x > 1.5 \)[/tex]:
[tex]\( f(x) \)[/tex] is negative again, and the parabola opens upwards and is positive.
- [tex]\( y \)[/tex] is increasing again because the absolute value will ensure a positive trend.
Synthesizing all this information gives us the following intervals:
- Increasing: [tex]\((- \infty, -2) \cup (1.5, \infty)\)[/tex]
- Decreasing: [tex]\((-2, -\frac{1}{4}) \cup (-\frac{1}{4}, 1.5)\)[/tex]
Thus, the correct answer from the provided choices is:
- Increasing for [tex]\( x < -2 \)[/tex] and [tex]\( x > \frac{3}{2} \)[/tex]; Decreasing for [tex]\( x > -2 \)[/tex] and [tex]\( 0 < x < \frac{3}{2} \)[/tex].
