To solve the problem, we need to find the absolute maximum and minimum values of the function \( f(x) = |2x^2 + 3x - 2| + \sin x \cos x \) on the interval \([0, 1]\).
### Step 1: Analyze the function inside the absolute value
First, we need to analyze the quadratic expression \( 2x^2 + 3x - 2 \) to find its roots, as they will help us determine where the absolute value function changes.
The roots can be found using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
where \( a = 2, b = 3, c = -2 \).
Calculating the discriminant:
\[
D = b^2 - 4ac = 3^2 - 4 \cdot 2 \cdot (-2) = 9 + 16 = 25
\]
Thus, the roots are:
\[
x = \frac{-3 \pm 5}{4}
\]
Calculating the roots:
\[
x_1 = \frac{2}{4} = \frac{1}{2}, \quad x_2 = \frac{-8}{4} = -2
\]
### Step 2: Determine the intervals
The roots indicate that \( 2x^2 + 3x - 2 \) changes sign at \( x = \frac{1}{2} \). We will check the sign of \( 2x^2 + 3x - 2 \) in the intervals \([0, \frac{1}{2})\) and \((\frac{1}{2}, 1]\).
- For \( x \in [0, \frac{1}{2}) \):
- Choose \( x = 0 \): \( 2(0)^2 + 3(0) - 2 = -2 \) (negative)
- For \( x = \frac{1}{2} \):
- \( 2(\frac{1}{2})^2 + 3(\frac{1}{2}) - 2 = 0 \) (zero)
- For \( x \in (\frac{1}{2}, 1] \):
- Choose \( x = 1 \): \( 2(1)^2 + 3(1) - 2 = 3 \) (positive)
Thus, we can conclude:
- \( 2x^2 + 3x - 2 < 0 \) for \( x \in [0, \frac{1}{2}) \)
- \( 2x^2 + 3x - 2 = 0 \) at \( x = \frac{1}{2} \)
- \( 2x^2 + 3x - 2 > 0 \) for \( x \in (\frac{1}{2}, 1] \)
### Step 3: Rewrite the function
Now we can rewrite the function \( f(x) \):
\[
f(x) =
\begin{cases}
-(2x^2 + 3x - 2) + \sin x \cos x & \text{if } x \in [0, \frac{1}{2}) \\
0 + \sin x \cos x & \text{if } x = \frac{1}{2} \\
(2x^2 + 3x - 2) + \sin x \cos x & \text{if } x \in (\frac{1}{2}, 1]
\end{cases}
\]
### Step 4: Find critical points
Next, we need to find the critical points of \( f(x) \) by differentiating \( f(x) \) and setting the derivative to zero.
1. For \( x \in [0, \frac{1}{2}) \):
\[
f'(x) = - (4x + 3) + \cos(2x)
\]
2. For \( x \in (\frac{1}{2}, 1] \):
\[
f'(x) = 4x + 3 + \cos(2x)
\]
### Step 5: Evaluate at critical points and endpoints
We evaluate \( f(x) \) at the endpoints \( x = 0, \frac{1}{2}, 1 \) and any critical points found in the previous step.
- At \( x = 0 \):
\[
f(0) = |2(0)^2 + 3(0) - 2| + \sin(0)\cos(0) = 2 + 0 = 2
\]
- At \( x = \frac{1}{2} \):
\[
f\left(\frac{1}{2}\right) = 0 + \sin\left(\frac{1}{2}\right)\cos\left(\frac{1}{2}\right) = \frac{1}{2}\sin(1)
\]
- At \( x = 1 \):
\[
f(1) = |2(1)^2 + 3(1) - 2| + \sin(1)\cos(1) = 3 + \frac{1}{2}\sin(2)
\]
### Step 6: Compare values
Now we compare the values:
- \( f(0) = 2 \)
- \( f\left(\frac{1}{2}\right) = \frac{1}{2}\sin(1) \)
- \( f(1) = 3 + \frac{1}{2}\sin(2) \)
### Step 7: Determine maximum and minimum
From the evaluations:
- The absolute minimum value is \( f\left(\frac{1}{2}\right) \).
- The absolute maximum value is \( f(1) \).
### Step 8: Calculate the sum
Finally, we find the sum of the absolute maximum and minimum values:
\[
\text{Sum} = f(1) + f\left(\frac{1}{2}\right) = \left(3 + \frac{1}{2}\sin(2)\right) + \frac{1}{2}\sin(1)
\]
### Final Answer
The sum of the absolute maximum and absolute minimum values of the function \( f(x) \) in the interval \([0, 1]\) is:
\[
\text{Sum} = 3 + \frac{1}{2}(\sin(2) + \sin(1))
\]
