Find the absolute maximum and minimum values of each of the following in the given intervals :
`f(x)=x^(3)-(5)/(2)x^(2)-2x+1, 0 le x le3.`

Answer

To find the absolute maximum and minimum values of the function \( f(x) = x^3 - \frac{5}{2}x^2 - 2x + 1 \) in the interval \( [0, 3] \), we will follow these steps:

### Step 1: Find the derivative of the function
We start by differentiating the function \( f(x) \).

\[ f'(x) = \frac{d}{dx}(x^3) - \frac{d}{dx}\left(\frac{5}{2}x^2\right) - \frac{d}{dx}(2x) + \frac{d}{dx}(1) \]

Calculating the derivatives, we get:

\[ f'(x) = 3x^2 - 5x - 2 \]

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

\[ 3x^2 - 5x - 2 = 0 \]

Now we can solve this quadratic equation using the factorization method or the quadratic formula. Here, we will factor it:

\[ 3x^2 - 6x + x - 2 = 0 \]

Grouping the terms:

\[ 3x(x - 2) + 1(x - 2) = 0 \]

Factoring out \( (x - 2) \):

\[ (3x + 1)(x - 2) = 0 \]

Setting each factor to zero gives us:

\[ 3x + 1 = 0 \quad \Rightarrow \quad x = -\frac{1}{3} \quad \text{(not in the interval [0, 3])} \]
\[ x - 2 = 0 \quad \Rightarrow \quad x = 2 \quad \text{(in the interval [0, 3])} \]

### Step 3: Evaluate the function at critical points and endpoints
We need to evaluate \( f(x) \) at the critical point \( x = 2 \) and the endpoints \( x = 0 \) and \( x = 3 \).

1. **At \( x = 0 \)**:
\[ f(0) = 0^3 - \frac{5}{2}(0^2) - 2(0) + 1 = 1 \]

2. **At \( x = 2 \)**:
\[ f(2) = 2^3 - \frac{5}{2}(2^2) - 2(2) + 1 \]
\[ = 8 - \frac{5}{2}(4) - 4 + 1 \]
\[ = 8 - 10 - 4 + 1 = -5 \]

3. **At \( x = 3 \)**:
\[ f(3) = 3^3 - \frac{5}{2}(3^2) - 2(3) + 1 \]
\[ = 27 - \frac{5}{2}(9) - 6 + 1 \]
\[ = 27 - \frac{45}{2} - 6 + 1 \]
\[ = 27 - 22.5 - 6 + 1 = -0.5 \]

### Step 4: Compare the values
Now we compare the function values at the endpoints and the critical point:

- \( f(0) = 1 \)
- \( f(2) = -5 \)
- \( f(3) = -0.5 \)

### Conclusion
The absolute maximum value of \( f(x) \) in the interval \( [0, 3] \) is \( 1 \) at \( x = 0 \), and the absolute minimum value is \( -5 \) at \( x = 2 \).

### Final Answer
- Absolute Maximum: \( 1 \) at \( x = 0 \)
- Absolute Minimum: \( -5 \) at \( x = 2 \)