Let `f(x)=(-1)^([x^(3)])`, where [.] denotest the greatest integer function. Then,

Options

A. f(x) is discontinuous at x`=n^(1//3), n in Z`
B. f(3/2)=1
C. `f'(0)=0"for all "x in (-1,1)`
D. none of these

Explanation

To solve the problem, we need to analyze the function \( f(x) = (-1)^{[\sqrt[3]{x}]} \), where \([.]\) denotes the greatest integer function (also known as the floor function). We will determine the points of discontinuity and evaluate the function at specific values.

### Step 1: Identify points of discontinuity
The greatest integer function \([\cdot]\) is discontinuous at integer values. Therefore, we need to find when \(\sqrt[3]{x}\) is an integer.

1. Set \(\sqrt[3]{x} = n\), where \(n\) is an integer.
2. This implies \(x = n^3\).

Thus, the function \(f(x)\) is discontinuous at \(x = n^3\) for all integers \(n\).

### Step 2: Write the points of discontinuity
The points of discontinuity for \(f(x)\) are at \(x = n^3\) where \(n \in \mathbb{Z}\).

### Step 3: Evaluate \(f\left(\frac{3}{2}\right)\)
Now, we need to evaluate \(f\left(\frac{3}{2}\right)\):

1. Calculate \(\left(\frac{3}{2}\right)^3 = \frac{27}{8} = 3.375\).
2. Now apply the greatest integer function: \([\sqrt[3]{3.375}] = [3.375] = 3\).
3. Therefore, \(f\left(\frac{3}{2}\right) = (-1)^3 = -1\).

### Step 4: Analyze the derivative \(f'(x)\)
Since \(f(x)\) is discontinuous at \(x = n^3\), we cannot find \(f'(x)\) at these points. Thus, \(f'(x)\) does not exist at \(x = n^3\).

### Conclusion
From the analysis, we conclude:
- The function \(f(x)\) is discontinuous at \(x = n^3\) for all integers \(n\).
- \(f\left(\frac{3}{2}\right) = -1\).
- The derivative \(f'(x)\) does not exist at \(x = n^3\).

### Final Answer
The correct option is that \(f(x)\) is discontinuous at \(x = n^3\) for \(n \in \mathbb{Z}\).

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