Number of points of discontinuity of the function `f(x) = [x^(1/x)], x > 0`, where [.] represents GIF is

Answer

To find the number of points of discontinuity of the function \( f(x) = \lfloor x^{1/x} \rfloor \) for \( x > 0 \), where \( \lfloor . \rfloor \) represents the greatest integer function (GIF), we will follow these steps:

### Step 1: Analyze the function \( f(x) = x^{1/x} \)

The function \( x^{1/x} \) is defined for \( x > 0 \). We need to understand its behavior to identify where the greatest integer function might cause discontinuities.

### Step 2: Find the derivative of \( x^{1/x} \)

To find the critical points, we can differentiate \( y = x^{1/x} \):
\[ y = e^{\frac{\ln x}{x}} \]
Taking the derivative using the chain rule:
\[ \frac{dy}{dx} = y \cdot \left( \frac{1 - \ln x}{x^2} \right) \]
Setting the derivative equal to zero gives us the critical points:
\[ 1 - \ln x = 0 \implies \ln x = 1 \implies x = e \]

### Step 3: Evaluate the function at critical points and endpoints

We will evaluate \( f(x) \) at \( x = 1 \), \( x = e \), and some values around these points to understand the behavior of \( x^{1/x} \).

- At \( x = 1 \):
\[ f(1) = \lfloor 1^{1/1} \rfloor = \lfloor 1 \rfloor = 1 \]

- At \( x = e \):
\[ f(e) = \lfloor e^{1/e} \rfloor \approx \lfloor 1.445 \rfloor = 1 \]

### Step 4: Check values around \( x = 1 \) and \( x = e \)

We need to check values slightly less than and greater than \( x = 1 \) and \( x = e \):

- For \( x < 1 \) (e.g., \( x = 0.5 \)):
\[ f(0.5) = \lfloor (0.5)^{2} \rfloor = \lfloor 0.25 \rfloor = 0 \]

- For \( x > 1 \) (e.g., \( x = 2 \)):
\[ f(2) = \lfloor (2)^{1/2} \rfloor = \lfloor \sqrt{2} \rfloor = 1 \]

### Step 5: Identify points of discontinuity

The function \( f(x) \) changes from 0 to 1 at \( x = 1 \). Since \( f(1) = 1 \) and \( f(0.5) = 0 \), there is a discontinuity at \( x = 1 \).

### Conclusion

The only point of discontinuity identified is at \( x = 1 \). Therefore, the number of points of discontinuity of the function \( f(x) = \lfloor x^{1/x} \rfloor \) for \( x > 0 \) is:

\[ \text{Number of points of discontinuity} = 1 \]