Discussion
If \(f(x) = |x^{2} + x - 6|\) is not differentiable at \(x = a\) and \(x = b\), then \(a^{2} + b^{2} =\)
Answer: D
\(a^{2} + b^{2} = \) 13.
The modulus of a polynomial has a corner where the polynomial crosses zero with a simple root. Here
\[ x^{2} + x - 6 = (x + 3)(x - 2) \]
so \(f\) is not differentiable at \(x = -3\) and \(x = 2\), giving \(a^{2} + b^{2} = 9 + 4 = 13\). Using the roots 3 and -2 (the wrong signs) gives the same 13, but 11, 12, 14 and 16 do not come from \(\{2, -3\}\).
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