Question

If the function given by \( f(x) = \left( \frac{4x + 1}{1 - 4x} \right)^{\frac{1}{3}} \) for \( x \neq 0 \) is continuous at \( x = 0 \), the value of \( f(0) \) is

• A. \( e^8 \)
• B. \( e^{10} \)
• C. \( e^{-8} \)
• D. \( e^{-10} \)

Hint:

When checking for continuity, always calculate the limit and ensure it matches the value of the function at the point of interest.

Answer 1

The Correct Option is A

Step 1: Check continuity at \( x = 0 \).
To ensure the function is continuous at \( x = 0 \), the limit of \( f(x) \) as \( x \) approaches 0 must equal \( f(0) \). Let's calculate the limit: \[ \lim_{x \to 0} \left( \frac{4x + 1}{1 - 4x} \right)^{\frac{1}{3}} \]
Step 2: Calculate the limit.
By substituting \( x = 0 \), we get: \[ \left( \frac{4(0) + 1}{1 - 4(0)} \right)^{\frac{1}{3}} = \left( \frac{1}{1} \right)^{\frac{1}{3}} = 1 \] Thus, \( f(0) = 1 \).
Step 3: Conclusion.
The value of \( f(0) \) is \( e^8 \).