At x = 0 , `f(x) {:(=x-2", if " x lt 0 ),(= 1//2", if " x = 0 ),(=x^(2)", if " x gt 0 ):}}` is

Options

A. continuous
B. discontinuous
C. not defined
D. undefined

Answer

To determine the continuity of the function \( f(x) \) at \( x = 0 \), we need to analyze the given piecewise function:

\[ f(x) = \begin{cases} x - 2 & \text{if } x < 0 \\ \frac{1}{2} & \text{if } x = 0 \\ x^2 & \text{if } x > 0 \end{cases} \]

### Step 1: Find \( f(0) \)

From the definition of the function, we have:

\[ f(0) = \frac{1}{2} \]

**Hint:** Identify the value of the function at the point of interest, which is \( x = 0 \).

### Step 2: Calculate the left-hand limit as \( x \) approaches 0

We need to find the limit of \( f(x) \) as \( x \) approaches 0 from the left (negative side):

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (x - 2) \]

Substituting \( x = 0 \):

\[ \lim_{x \to 0^-} f(x) = 0 - 2 = -2 \]

**Hint:** When calculating the left-hand limit, use the expression that applies for \( x < 0 \).

### Step 3: Calculate the right-hand limit as \( x \) approaches 0

Now, we find the limit of \( f(x) \) as \( x \) approaches 0 from the right (positive side):

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x^2) \]

Substituting \( x = 0 \):

\[ \lim_{x \to 0^+} f(x) = 0^2 = 0 \]

**Hint:** For the right-hand limit, use the expression that applies for \( x > 0 \).

### Step 4: Compare the limits and the function value

We now compare the left-hand limit, right-hand limit, and the function value at \( x = 0 \):

- Left-hand limit: \( -2 \)
- Right-hand limit: \( 0 \)
- Function value: \( f(0) = \frac{1}{2} \)

Since:

\[ \lim_{x \to 0^-} f(x) = -2 \quad \text{and} \quad \lim_{x \to 0^+} f(x) = 0 \quad \text{and} \quad f(0) = \frac{1}{2} \]

These values are not equal:

\[ -2 \neq \frac{1}{2} \quad \text{and} \quad 0 \neq \frac{1}{2} \]

### Conclusion

Since the left-hand limit, right-hand limit, and the function value at \( x = 0 \) are not equal, the function \( f(x) \) is discontinuous at \( x = 0 \).

**Final Answer:** The function is discontinuous at \( x = 0 \).