Question

Let \( f : \mathbb{R} \setminus \{0\} \to \mathbb{R} \) be defined by \( f(x) = x + \dfrac{1}{x^3} \). 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: Find the derivative

$$f'(x) = 1 - \frac{3}{x^4}$$

Step 2: Determine where $f'(x) = 0$

$$1 - \frac{3}{x^4} = 0$$ $$x^4 = 3$$ $$x = \pm \sqrt[4]{3}$$

Let $\alpha = \sqrt[4]{3} \approx 1.316$

Critical points are at $x = \alpha$ and $x = -\alpha$.

Step 3: Analyze the sign of $f'(x)$

$$f'(x) = 1 - \frac{3}{x^4} = \frac{x^4 - 3}{x^4}$$

Since $x^4 > 0$ for all $x \neq 0$, the sign of $f'(x)$ depends on $(x^4 - 3)$:

• $f'(x) > 0$ when $x^4 > 3$, i.e., when $|x| > \sqrt[4]{3}$
• $f'(x) < 0$ when $x^4 < 3$, i.e., when $|x| < \sqrt[4]{3}$

Intervals:

• On $(-\infty, -\alpha)$: $f'(x) > 0$ → $f$ is increasing
• On $(-\alpha, 0)$: $f'(x) < 0$ → $f$ is decreasing
• On $(0, \alpha)$: $f'(x) < 0$ → $f$ is decreasing
• On $(\alpha, \infty)$: $f'(x) > 0$ → $f$ is increasing

Step 4: Check each option

(A) $(-\infty, -1)$:

Since $-1 > -\alpha$ (because $\alpha \approx 1.316$), the interval $(-\infty, -1)$ includes parts where $f$ is increasing (for $x < -\alpha$) and parts where $f$ is decreasing (for $-\alpha < x < -1$).

Therefore, $f$ is not one-one on $(-\infty, -1)$. 

(B) $(0, 1)$:

Since $1 < \alpha \approx 1.316$, the entire interval $(0, 1) \subset (0, \alpha)$.

On $(0, \alpha)$, we have $f'(x) < 0$, so $f$ is strictly decreasing.

Therefore, $f$ is one-one on $(0, 1)$. 

(C) $(0, 2)$:

Since $\alpha \approx 1.316 \in (0, 2)$, the interval $(0, 2)$ contains the critical point $\alpha$.

On $(0, \alpha)$, $f$ is decreasing, and on $(\alpha, 2)$, $f$ is increasing.

Therefore, $f$ is not one-one on $(0, 2)$. 

(D) $(0, \infty)$:

This interval contains the critical point $\alpha$ where $f$ changes from decreasing to increasing.

Therefore, $f$ is not one-one on $(0, \infty)$. 

Step 5: Verify option (B) explicitly

For $(0, 1)$: Since $f'(x) < 0$ throughout this interval, $f$ is strictly monotonic (decreasing), which guarantees injectivity.

If $x_1, x_2 \in (0, 1)$ with $x_1 < x_2$, then $f(x_1) > f(x_2)$, so $f(x_1) \neq f(x_2)$.

Answer: (B)