Discussion
Consider the function \(f(x)=|x|\) in the interval \(-1 \le x \le 1\). At the point x = 0, f(x) is
Answer: C
At x = 0, \(|x|\) is continuous and non-differentiable.
It is continuous because both one-sided limits equal the value of the function: \(\lim_{x \to 0}|x| = 0 = f(0)\).
It is not differentiable because the two one-sided derivatives differ:
\[ \begin{aligned} f'(0^{+}) &= \lim_{h \to 0^{+}}\frac{h}{h} = 1 \\ f'(0^{-}) &= \lim_{h \to 0^{-}}\frac{-h}{h} = -1 \end{aligned} \]
The graph has a sharp corner at the origin. 'Non-continuous and differentiable' is impossible, since differentiability implies continuity, and the function is plainly continuous, which rules out the remaining two options.
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