Discussion
If \(f(x) = \frac{|x|}{x^{2}}\) , then \(f'(2)\) is equal to
Answer: D
\(f'(2)\) is \(-\frac{1}{4}\).
Near x = 2, x is positive, so \(|x| = x\) and \(f(x) = \frac{x}{x^{2}} = \frac{1}{x}\).
\[ f'(x) = -\frac{1}{x^{2}} \Rightarrow f'(2) = -\frac{1}{4} \]
\(\frac{1}{4}\) drops the minus sign, and \(-\frac{1}{2}\) is \(-\frac{1}{x}\) instead of \(-\frac{1}{x^{2}}\). The modulus matters only for negative x, where \(f(x) = -\frac{1}{x}\) and the derivative would be \(+\frac{1}{x^{2}}\); at x = 2 that branch is not used.
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