The function `f(x) = x^(3)/4 - sin pi x + 3 " on " [-2,2]` takes the value

Options

A. 1
B. `16//3`
C. 6
D. 8

Answer

To solve the problem, we need to determine the values that the function \( f(x) = \frac{x^3}{4} - \sin(\pi x) + 3 \) takes on the interval \([-2, 2]\). We will use the Intermediate Value Theorem (IVT) for this purpose.

### Step-by-Step Solution:

1. **Define the Function**:
The function is given as:
\[ f(x) = \frac{x^3}{4} - \sin(\pi x) + 3 \]

2. **Check Continuity**:
We need to check if the function is continuous on the interval \([-2, 2]\).
- The term \(\frac{x^3}{4}\) is a polynomial and is continuous everywhere.
- The term \(-\sin(\pi x)\) is also continuous everywhere since sine is a continuous function.
- The constant \(3\) is continuous.
Therefore, \(f(x)\) is continuous on \([-2, 2]\).

3. **Evaluate the Function at the Endpoints**:
Now we will evaluate \(f(x)\) at the endpoints of the interval:
- Calculate \(f(-2)\):
\[ f(-2) = \frac{(-2)^3}{4} - \sin(-2\pi) + 3 = \frac{-8}{4} - 0 + 3 = -2 + 3 = 1 \]
- Calculate \(f(2)\):
\[ f(2) = \frac{(2)^3}{4} - \sin(2\pi) + 3 = \frac{8}{4} - 0 + 3 = 2 + 3 = 5 \]

4. **Apply the Intermediate Value Theorem**:
Since \(f(x)\) is continuous on \([-2, 2]\) and we have:
\[ f(-2) = 1 \quad \text{and} \quad f(2) = 5 \]
By the Intermediate Value Theorem, \(f(x)\) takes all values between \(f(-2)\) and \(f(2)\), which means \(f(x)\) takes all values in the interval \([1, 5]\).

5. **Determine Possible Values**:
We need to find which of the given options lies between \(1\) and \(5\). The options are:
- \(1\)
- \(5\)
- \(16/3\)
- \(6\)

From our analysis:
- \(1\) is included.
- \(5\) is included.
- \(16/3 \approx 5.33\) which is greater than \(5\).
- \(6\) is also greater than \(5\).

Therefore, the only values that \(f(x)\) can take from the options provided are \(1\) and \(5\).

### Conclusion:
The function \(f(x)\) takes the value \(1\) and \(5\) on the interval \([-2, 2]\).