Consider the function `f(x) = |x^(3) + 1|`. Then,

Options

A. domain of f `x in R`
B. range of f is `R^(+)`
C. f has no inverse
D. f is continuous and differentiable for every `x in R`

Answer

To solve the problem regarding the function \( f(x) = |x^3 + 1| \), we will analyze the function step by step.

### Step 1: Understanding the Function
The function \( f(x) = |x^3 + 1| \) involves the absolute value of the polynomial \( x^3 + 1 \). To understand how the absolute value affects the function, we first need to analyze the polynomial \( x^3 + 1 \).

### Step 2: Finding the Roots of the Polynomial
To find where the polynomial \( x^3 + 1 \) changes sign, we set it equal to zero:
\[ x^3 + 1 = 0 \implies x^3 = -1 \implies x = -1 \]
Thus, the polynomial \( x^3 + 1 \) changes sign at \( x = -1 \).

### Step 3: Analyzing the Sign of \( x^3 + 1 \)
- For \( x < -1 \): \( x^3 + 1 < 0 \) (the function is negative)
- For \( x = -1 \): \( x^3 + 1 = 0 \) (the function is zero)
- For \( x > -1 \): \( x^3 + 1 > 0 \) (the function is positive)

### Step 4: Expressing \( f(x) \) in Piecewise Form
Based on the sign analysis:
\[ f(x) = \begin{cases} -(x^3 + 1) & \text{if } x < -1 \\ 0 & \text{if } x = -1 \\ x^3 + 1 & \text{if } x > -1 \end{cases} \]

### Step 5: Finding the Domain of \( f(x) \)
The domain of \( f(x) \) is the set of all real numbers since there are no restrictions on \( x \):
\[ \text{Domain: } \mathbb{R} \]

### Step 6: Finding the Range of \( f(x) \)
- For \( x < -1 \): \( f(x) = -x^3 - 1 \) which approaches \( \infty \) as \( x \) approaches \( -\infty \) and approaches \( 0 \) as \( x \) approaches \( -1 \).
- For \( x = -1 \): \( f(-1) = 0 \).
- For \( x > -1 \): \( f(x) = x^3 + 1 \) which approaches \( 1 \) as \( x \) approaches \( -1 \) and approaches \( \infty \) as \( x \) approaches \( \infty \).

Thus, the range of \( f(x) \) is:
\[ \text{Range: } [0, \infty) \]

### Step 7: Continuity of \( f(x) \)
The function \( f(x) \) is continuous everywhere since both pieces of the piecewise function are continuous, and they meet at \( x = -1 \) where \( f(-1) = 0 \).

### Step 8: Differentiability of \( f(x) \)
To check for differentiability, we need to consider the point where \( x = -1 \):
- For \( x < -1 \): The derivative \( f'(x) = -3x^2 \).
- For \( x > -1 \): The derivative \( f'(x) = 3x^2 \).

At \( x = -1 \):
- The left-hand derivative is \( f'(-1^-) = -3(-1)^2 = -3 \).
- The right-hand derivative is \( f'(-1^+) = 3(-1)^2 = 3 \).

Since the left-hand and right-hand derivatives are not equal, \( f(x) \) is not differentiable at \( x = -1 \).

### Conclusion
- **Domain**: \( \mathbb{R} \)
- **Range**: \( [0, \infty) \)
- **Continuity**: \( f(x) \) is continuous for all \( x \in \mathbb{R} \).
- **Differentiability**: \( f(x) \) is not differentiable at \( x = -1 \).