If f(x) = `{{:(0, "for" x=0),(x-3, "for" x gt0):}` then function f(x) is

Options

A. increasing when `x gt 0`
B. strictly increasing when x`gt` 0
C. strictly increasing at x =0
D. not continuous at x =0 and so it is not increasing when x`gt0`

Answer

To analyze the function \( f(x) \) given by:

\[ f(x) = \begin{cases} 0 & \text{for } x = 0 \\ x - 3 & \text{for } x > 0 \end{cases} \]

we need to determine the behavior of this function, particularly regarding its continuity and whether it is increasing or strictly increasing.

### Step 1: Check the continuity at \( x = 0 \)

To check for continuity at \( x = 0 \), we need to evaluate:

1. \( f(0) \)
2. \( \lim_{x \to 0^+} f(x) \)
3. \( \lim_{x \to 0^-} f(x) \)

Calculating these:

- \( f(0) = 0 \)
- For \( x > 0 \), \( f(x) = x - 3 \). Therefore, \( \lim_{x \to 0^+} f(x) = 0 - 3 = -3 \).
- Since there is no definition for \( f(x) \) when \( x < 0 \), \( \lim_{x \to 0^-} f(x) \) does not exist.

Since \( \lim_{x \to 0^+} f(x) \neq f(0) \), the function is **not continuous at \( x = 0 \)**.

### Step 2: Analyze the function for \( x > 0 \)

For \( x > 0 \), \( f(x) = x - 3 \).

- The derivative \( f'(x) \) can be calculated as follows:

\[ f'(x) = \frac{d}{dx}(x - 3) = 1 \]

Since \( f'(x) = 1 > 0 \) for all \( x > 0 \), this indicates that the function is **increasing for \( x > 0 \)**.

### Step 3: Determine if the function is strictly increasing

A function is strictly increasing if \( f(x_1) < f(x_2) \) for any \( x_1 < x_2 \).

- For \( x_1 < x_2 \) in the interval \( (0, \infty) \):

\[ f(x_1) = x_1 - 3 \quad \text{and} \quad f(x_2) = x_2 - 3 \]

Since \( x_1 < x_2 \), it follows that:

\[ x_1 - 3 < x_2 - 3 \]

Thus, \( f(x_1) < f(x_2) \), confirming that \( f(x) \) is **strictly increasing** for \( x > 3 \) but not for \( 0 < x < 3 \) since \( f(x) \) takes negative values in that interval.

### Conclusion

1. The function is **not continuous at \( x = 0 \)**.
2. The function is **increasing for \( x > 0 \)** but **not strictly increasing** for \( 0 < x < 3 \) since it takes negative values.

### Final Answer

The function \( f(x) \) is:
- Not continuous at \( x = 0 \)
- Increasing for \( x > 0 \)
- Not strictly increasing for \( x > 0 \)