Let f(x) = x|x| . Then f(x) is:

The correct answer is Option (3) → Both Continuous and Differentiable at \(x = 0\)

Writing the function without the modulus:

    \( f(x) = \begin{cases} x^2, & x \ge 0\\ -x^2, & x \lt 0 \end{cases} \)

Continuity at 0: both one-sided limits are 0 and f(0) = 0.

    \( \displaystyle \lim_{x\to 0^{+}} x^2 = 0,\quad \lim_{x\to 0^{-}} \left(-x^2\right) = 0 \)

So f is continuous at 0.

Differentiability at 0:

    \( \displaystyle \lim_{h\to 0}\frac{f(0+h)-f(0)}{h} = \lim_{h\to 0}\frac{h|h|}{h} \)

    \( \displaystyle = \lim_{h\to 0}|h| = 0 \)

The limit exists from both sides (the left and right derivatives are both 0), so f is differentiable at 0, with f′(0) = 0.

Hence f is both continuous and differentiable at x = 0.

Tip: |x| has a corner at 0, but \(x|x|\) does not: the extra factor x makes the left and right slopes both 0.