Limits and Continuity
On what intervals are the following functions continuous?


c. h(x) = x⁻²/³

Verified step by step guidance

1. Identify the type of function: The function given is \( h(x) = x^{-2/3} \), which can be rewritten as \( h(x) = \frac{1}{x^{2/3}} \). This is a rational function where the denominator is \( x^{2/3} \).
2. Determine where the function is undefined: A rational function is undefined where its denominator is zero. Set \( x^{2/3} = 0 \) and solve for \( x \). This gives \( x = 0 \).
3. Analyze the continuity: A function is continuous on an interval if it is defined and does not have any breaks, holes, or asymptotes within that interval. Since \( h(x) \) is undefined at \( x = 0 \), it cannot be continuous there.
4. Identify the intervals of continuity: Since the function is only undefined at \( x = 0 \), it is continuous on the intervals \((-\infty, 0)\) and \((0, \infty)\).
5. Conclude the intervals of continuity: The function \( h(x) = x^{-2/3} \) is continuous on the intervals \((-\infty, 0)\) and \((0, \infty)\), but not at \( x = 0 \).