Which one of the following is correct in respect of the function `f(x)=|x|+x^(2)`

Options

A. f(x) is not continuous at x=0
B. f(x) is differentiable at x=0
C. f(x) us continuous but not differentible at x=0
D. None of the above

Explanation

To determine the properties of the function \( f(x) = |x| + x^2 \), we will check its continuity and differentiability at \( x = 0 \).

### Step 1: Define the function based on the value of \( x \)

The function \( f(x) \) can be expressed differently depending on whether \( x \) is non-negative or negative:

- For \( x \geq 0 \):
\[ f(x) = |x| + x^2 = x + x^2 \]

- For \( x < 0 \):
\[ f(x) = |x| + x^2 = -x + x^2 \]

### Step 2: Check continuity at \( x = 0 \)

To check continuity at \( x = 0 \), we need to evaluate the left-hand limit, right-hand limit, and the function value at \( x = 0 \).

1. **Left-hand limit** as \( x \) approaches 0 from the left (\( x \to 0^- \)):
\[ f(0 - h) = -h + (-h)^2 = -h + h^2 \]
Taking the limit as \( h \to 0 \):
\[ \lim_{h \to 0} (-h + h^2) = 0 \]

2. **Right-hand limit** as \( x \) approaches 0 from the right (\( x \to 0^+ \)):
\[ f(0 + h) = h + h^2 \]
Taking the limit as \( h \to 0 \):
\[ \lim_{h \to 0} (h + h^2) = 0 \]

3. **Function value** at \( x = 0 \):
\[ f(0) = |0| + 0^2 = 0 \]

Since both limits and the function value at \( x = 0 \) are equal:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) = 0 \]
Thus, \( f(x) \) is continuous at \( x = 0 \).

### Step 3: Check differentiability at \( x = 0 \)

To check differentiability, we will find the left-hand derivative and right-hand derivative at \( x = 0 \).

1. **Left-hand derivative**:
\[ f'(0^-) = \lim_{h \to 0} \frac{f(0 - h) - f(0)}{-h} = \lim_{h \to 0} \frac{-h + h^2 - 0}{-h} = \lim_{h \to 0} \frac{-h + h^2}{-h} \]
Simplifying:
\[ = \lim_{h \to 0} (1 - h) = 1 \]

2. **Right-hand derivative**:
\[ f'(0^+) = \lim_{h \to 0} \frac{f(0 + h) - f(0)}{h} = \lim_{h \to 0} \frac{h + h^2 - 0}{h} = \lim_{h \to 0} (1 + h) = 1 \]

Since the left-hand derivative and right-hand derivative are not equal:
\[ f'(0^-) = 1 \quad \text{and} \quad f'(0^+) = 1 \]
Thus, \( f(x) \) is differentiable at \( x = 0 \).

### Conclusion

The function \( f(x) = |x| + x^2 \) is continuous at \( x = 0 \) and differentiable at \( x = 0 \). The correct option is that the function is continuous but not differentiable at \( x = 0 \).

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