Discussion
Let \(h: \mathbb{R} \to \mathbb{R}\) and \(g: \mathbb{R} \to \mathbb{R}\) be two bounded functions. Define the function \(f: \mathbb{R} \to \mathbb{R}\) by
\[ f(x) = \begin{cases} x g(x) + x^{2} h(x), & x \gt 0, \\ cx + dx^{2}, & x \leq 0, \end{cases} \]
where \(c, d\) are real constants. Consider the following statements.
P: If \(c = g(0) = h(0)\), then \(f\) is differentiable at \(x = 0\).
Q: If \(f, g\), and \(h\) are thrice differentiable functions with \(g(0) = 0\) and \(g'(0) + h(0) \gt 0\), then \(f\) has a local minimum at \(x = 0\).
Then
Answer: B
P is NOT correct and Q is correct.
P: \(f(0) = 0\). The left derivative at 0 is \(c\). The right derivative is
\[ \begin{aligned} &\lim_{x \to 0^{+}} \frac{x g(x) + x^{2} h(x)}{x} \\ &= \lim_{x \to 0^{+}} \left(g(x) + x h(x)\right) \end{aligned} \]
Because \(h\) is bounded, \(x h(x) \to 0\), but \(g\) is only bounded, not continuous, so \(g(x)\) need not tend to \(g(0)\) or to anything. Knowing \(c = g(0) = h(0)\) does not make the two one-sided derivatives agree, so P fails.
Q: now \(f, g, h\) are smooth. For \(x \gt 0\), \(f'(x) = g + xg' + 2xh + x^{2}h'\), so \(f'(0) = g(0) = 0\), and \(f''(0) = 2g'(0) + 2h(0) \gt 0\). Since \(f\) is twice differentiable at 0, the left side must match (\(c = 0\), \(2d = f''(0)\)). With \(f'(0) = 0\) and \(f''(0) \gt 0\), the second derivative test gives a local minimum.
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