Applications of Derivatives
Local Maxima and Minima
Grade 12
Question:
<p>Let <span class="math">\(f(x) = \begin{cases} x^3 + x^2 + 10x, & x < 0 \\ -3\sin x, & x \geq 0 \end{cases}\)</span>, then at <span class="math">\(x = 0\)</span>, <span class="math">\(f(x)\)</span> is</p>
<p>(a) local minimum</p>
<p>(b) local maximum</p>
<p>(c) Neither maximum nor minimum</p>
<p>(d) None of the above</p>
Step-by-Step Solution
Key Concept: When a function is not differentiable at a point, examine the left and right derivatives. If f'₋ > 0 and f'₊ < 0, the function has a local maximum.
<p><strong>Step 1:</strong> Clearly, <span class="math">$f(x)$</span> is continuous at <span class="math">$x = 0$</span> but not differentiable at <span class="math">$x = 0$</span> as <span class="math">$f(0^-) = f(0^+) = f(0) = 0$</span>.</p><p><strong>Step 2:</strong> Calculate left and right derivatives at <span class="math">$x = 0$</span>:</p><p><span class="math">$f'_-(0) = \lim_{h \to 0} \frac{-h^3 + h^2 - 10h - 0}{-h} = \lim_{h \to 0} \frac{-h^3 + h^2 - 10h}{-h} = 10$</span></p><p><span class="math">$f'_+(0) = \lim_{h \to 0} \frac{-3\sin h - 0}{h} = \lim_{h \to 0} \frac{-3\sin h}{h} = -3$</span></p><p><strong>Step 3:</strong> Since <span class="math">$f'_-(0) > 0$</span> and <span class="math">$f'_+(0) < 0$</span>, the derivative changes from positive to negative, indicating <span class="math">$x = 0$</span> is a point of local maximum.</p><p>∴ Answer is (b).</p>
Correct Answer: b