Find the absolute maximum value and absolute minimum value of the following question `f(x) =((1)/(2)-x)^(2)+x^(3)" in "[-2,2.5]`

Answer

To find the absolute maximum and minimum values of the function \( f(x) = \left(\frac{1}{2} - x\right)^2 + x^3 \) on the interval \([-2, 2.5]\), we will follow these steps:

### Step 1: Find the derivative of the function
First, we need to differentiate the function \( f(x) \).

\[ f(x) = \left(\frac{1}{2} - x\right)^2 + x^3 \]

Using the chain rule and power rule, we differentiate:

\[ f'(x) = 2\left(\frac{1}{2} - x\right)(-1) + 3x^2 \]

Simplifying this, we get:

\[ f'(x) = -2\left(\frac{1}{2} - x\right) + 3x^2 = -1 + 2x + 3x^2 \]

Thus,

\[ f'(x) = 3x^2 + 2x - 1 \]

### Step 2: Set the derivative equal to zero
To find critical points, we set the derivative equal to zero:

\[ 3x^2 + 2x - 1 = 0 \]

### Step 3: Solve the quadratic equation
We will use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 3, b = 2, c = -1 \).

Calculating the discriminant:

\[ b^2 - 4ac = 2^2 - 4 \cdot 3 \cdot (-1) = 4 + 12 = 16 \]

Now applying the quadratic formula:

\[ x = \frac{-2 \pm \sqrt{16}}{2 \cdot 3} = \frac{-2 \pm 4}{6} \]

This gives us two solutions:

\[ x_1 = \frac{2}{6} = \frac{1}{3}, \quad x_2 = \frac{-6}{6} = -1 \]

### Step 4: Evaluate the function at critical points and endpoints
Now we evaluate \( f(x) \) at the critical points and the endpoints of the interval \([-2, 2.5]\).

1. **At \( x = -2 \)**:
\[ f(-2) = \left(\frac{1}{2} - (-2)\right)^2 + (-2)^3 = \left(\frac{1}{2} + 2\right)^2 - 8 = \left(\frac{5}{2}\right)^2 - 8 = \frac{25}{4} - 8 = \frac{25}{4} - \frac{32}{4} = -\frac{7}{4} \]

2. **At \( x = -1 \)**:
\[ f(-1) = \left(\frac{1}{2} - (-1)\right)^2 + (-1)^3 = \left(\frac{1}{2} + 1\right)^2 - 1 = \left(\frac{3}{2}\right)^2 - 1 = \frac{9}{4} - 1 = \frac{9}{4} - \frac{4}{4} = \frac{5}{4} \]

3. **At \( x = \frac{1}{3} \)**:
\[ f\left(\frac{1}{3}\right) = \left(\frac{1}{2} - \frac{1}{3}\right)^2 + \left(\frac{1}{3}\right)^3 = \left(\frac{3}{6} - \frac{2}{6}\right)^2 + \frac{1}{27} = \left(\frac{1}{6}\right)^2 + \frac{1}{27} = \frac{1}{36} + \frac{1}{27} \]
Finding a common denominator (108):
\[ = \frac{3}{108} + \frac{4}{108} = \frac{7}{108} \]

4. **At \( x = 2.5 \)**:
\[ f(2.5) = \left(\frac{1}{2} - 2.5\right)^2 + (2.5)^3 = \left(-2\right)^2 + 15.625 = 4 + 15.625 = 19.625 = \frac{157}{8} \]

### Step 5: Compare values
Now we compare the values obtained:

- \( f(-2) = -\frac{7}{4} \)
- \( f(-1) = \frac{5}{4} \)
- \( f\left(\frac{1}{3}\right) = \frac{7}{108} \)
- \( f(2.5) = \frac{157}{8} \)

### Conclusion
The absolute maximum value is \( \frac{157}{8} \) and the absolute minimum value is \( -\frac{7}{4} \).