Question

If \( f(x) \) is given as: \[ f(x) = \begin{cases} \frac{8}{x^3} - 6x, & \text{if } 0 < x \leq 1 \\ \frac{x - 1}{\sqrt{x}-1}, & \text{if } x > 1 \end{cases} \] is a real-valued function, then at \( x = 1 \), \( f \) is:

• A. Continuous and differentiable
• B. Continuous but not differentiable
• C. Neither continuous nor differentiable
• D. Differentiable but not continuous

Hint:

To check differentiability, compute left-hand and right-hand derivatives separately.

Answer 1

The Correct Option is B

Solution:

A function \( f(x) \) is continuous at \( x = a \) if and only if

\[ \lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x) = f(a). \]

In this case,

\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \left( \frac{8}{x^3} - 6x \right) = \frac{8}{1^3} - 6 \cdot 1 = 8 - 6 = 2, \]

and

\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} \frac{x - 1}{\sqrt{x} - 1} = \lim_{x \to 1^+} \frac{(\sqrt{x} - 1)(\sqrt{x} + 1)}{\sqrt{x} - 1} = \lim_{x \to 1^+} (\sqrt{x} + 1) = \sqrt{1} + 1 = 2, \]

and \( f(1) = \frac{8}{1^3} - 6 \cdot 1 = 2 \), so \( f \) is continuous at \( x = 1 \).

For \( x \leq 1 \),

\[ f'(x) = - \frac{24}{x^4} - 6, \]

and for \( x > 1 \),

\[ f'(x) = \frac{(\sqrt{x} - 1) - (x - 1) \cdot \frac{1}{2 \sqrt{x}}}{(\sqrt{x} - 1)^2} = \frac{2x - 2 \sqrt{x} - x + \sqrt{x}}{2 \sqrt{x} (\sqrt{x} - 1)^2} = \frac{x - \sqrt{x}}{2 \sqrt{x} (\sqrt{x} - 1)^2} = \frac{\sqrt{x} (\sqrt{x} - 1)}{2 \sqrt{x} (\sqrt{x} - 1)^2} = \frac{1}{2 (\sqrt{x} - 1)}. \]