Applications of Derivatives
Local Maxima and Minima
Grade 12
Question:
<p>For \(f(x) = x^2 - 4|x|\) and \(g(x) = \begin{cases} \min\{f(t): -6 \leq t \leq x\}, & x \in [-6, 0] \\ \max\{f(t): 0 < t \leq x\}, & x \in (0, 6] \end{cases}\), \(g(x)\) has</p>
<p>(a) exactly one point of local minima</p>
<p>(b) exactly one point of local maxima</p>
<p>(c) no point of local maxima but exactly one point of local minima</p>
<p>(d) neither a point of local maxima nor minima</p>
Step-by-Step Solution
Key Concept: Analyze the piecewise function $g(x)$ by finding the min/max of $f(x)$ on appropriate intervals.
<p>For $x \in [-6, 0]$: $f(x) = x^2 - 4|x| = x^2 + 4x$. The minimum occurs at $x = -2$ with value $f(-2) = -4$.</p><p>So $g(x) = \begin{cases} f(x) & -6 \leq x \leq -2 \\ -4 & -2 < x \leq 0 \end{cases}$</p><p>For $x \in (0, 6]$: $f(x) = x^2 - 4x$. The minimum occurs at $x = 2$ with value $f(2) = -4$.</p><p>So $g(x) = \begin{cases} f(x) & 0 < x \leq 2 \\ -4 & 2 < x \leq 6 \end{cases}$</p><p>Thus $g$ has a local minimum at $x = 2$ and no local maximum.</p>
Correct Answer: C