Question

Let \( f : \mathbb{R} \setminus \{0\} \to \mathbb{R} \) be defined by \( f(x) = x + \frac{1{x^3} \). On which of the following interval(s) is \( f \) one-to-one?}

• A. \( (-\infty, -1) \)
• B. \( (0, 1) \)
• C. \( (0, 2) \)
• D. \( (0, \infty) \)

Hint:

To determine where a function is one-to-one, check the sign of its derivative. A function is one-to-one where the derivative does not change sign.

Answer 1

The Correct Option is B

Step 1: Analyzing the function.
The function \( f(x) = x + \frac{1}{x^3} \) involves both linear and rational terms. To determine where it is one-to-one, we need to analyze its derivative.

Step 2: Derivative of \( f(x) \).
We calculate the derivative: \[ f'(x) = 1 - \frac{3}{x^4} \] The function is increasing or decreasing based on the sign of \( f'(x) \).

Step 3: Interval of one-to-one behavior.
For \( f(x) \) to be one-to-one, \( f'(x) \) must not change sign. Solving \( f'(x) = 0 \) gives \( x = -1 \). On \( (-\infty, -1) \), \( f'(x) \) is positive, so \( f(x) \) is increasing and one-to-one. Thus, the correct answer is \( (-\infty, -1) \).

Step 4: Conclusion.
The correct answer is (A) \( (-\infty, -1) \).