Question

Let \( f : [0,1] \to \mathbb{R} \) be given by \[ f(x) = \frac{\left( \frac{1}{1+x^3} \right)^3 + \left( \frac{1}{1-x^3} \right)^3}{8(1+x)}. \] Then \( \max \{ f(x) : x \in [0,1] \} - \min \{ f(x) : x \in [0,1] \} \) is ...........

Hint:

When analyzing functions with denominators that approach zero, check the function's behavior at those points to determine if it is undefined or unbounded.

Answer 1

Correct Answer: 0.25

Step 1: Analyze the function.
We are given the function: \[ f(x) = \frac{\left( \frac{1}{1+x^3} \right)^3 + \left( \frac{1}{1-x^3} \right)^3}{8(1+x)}. \] To evaluate \( \max \{ f(x) : x \in [0,1] \} - \min \{ f(x) : x \in [0,1] \} \), we first need to understand the behavior of \( f(x) \) over the interval \( [0, 1] \).
Step 2: Check boundary points.
Let's evaluate the function at the endpoints of the interval: - At \( x = 0 \), the function becomes: \[ f(0) = \frac{\left( \frac{1}{1+0^3} \right)^3 + \left( \frac{1}{1-0^3} \right)^3}{8(1+0)} = \frac{2}{8} = \frac{1}{4}. \] - At \( x = 1 \), the function becomes: \[ f(1) = \frac{\left( \frac{1}{1+1^3} \right)^3 + \left( \frac{1}{1-1^3} \right)^3}{8(1+1)} = \frac{\left( \frac{1}{2} \right)^3 + \left( \frac{1}{0} \right)^3}{16}. \] However, the term \( \left( \frac{1}{0} \right)^3 \) leads to a division by zero, so \( f(1) \) is undefined.
Step 3: Analyze for critical points.
We observe that the function is undefined at \( x = 1 \), so we focus on evaluating the function in the open interval \( (0, 1) \). Based on the behavior at \( x = 0 \) and \( x = 1 \), and the fact that the function increases as \( x \) approaches 1 from the left, we find: \[ \max \{ f(x) : x \in [0,1] \} = f(0) = \frac{1}{4}, \quad \min \{ f(x) : x \in [0,1] \} \to \infty \text{ as } x \to 1. \] Thus, the value of \( \max f(x) - \min f(x) \) is unbounded, and we conclude: \[ \boxed{\infty}. \]