Applications of Derivatives
Extrema of Functions
Grade 12
Question:
<p>Let \(f(x) = \min\left(\frac{1}{2} - \frac{3x^2}{4}, \frac{5x^2}{4}\right)\) for \(0 \leq x \leq 1\). Then the maximum value of \(f(x)\) is:</p>
<p>(a) 0</p>
<p>(b) \(\frac{5}{64}\)</p>
<p>(c) \(\frac{5}{4}\)</p>
<p>(d) \(\frac{5}{16}\)</p>
Step-by-Step Solution
Key Concept: The maximum of a minimum function occurs at the intersection point of the two curves.
<p>Set $\frac{1}{2} - \frac{3x^2}{4} = \frac{5x^2}{4}$. Then $\frac{1}{2} = 2x^2$, so $x^2 = \frac{1}{4}$, thus $x = \frac{1}{2}$. At $x = \frac{1}{2}$, $f\left(\frac{1}{2}\right) = \frac{5}{4} \cdot \frac{1}{4} = \frac{5}{16}$.</p>
Correct Answer: d