Applications of Derivatives
Maxima and Minima
Grade 12
Question:
<p>Let \(f(x) = \begin{cases} \frac{1}{2}x^3 - x^2 + 10x + 5, & x \leq 1 \\ 4 - 2^x + \log_2(b^2 - 2), & x > 1 \end{cases}\)</p><p>The set of values of \(b\) for which \(f(x)\) has greatest value at \(x = 1\) is given by</p>
<p>(a) \(1 \leq b \leq 2\)</p>
<p>(b) \(b \in \{1, 2\}\)</p>
<p>(c) \(b \in (-\infty, -1)\)</p>
<p>(d) \([-\sqrt{30}, -2) \cup (2, \sqrt{30}]\)</p>
Step-by-Step Solution
Key Concept: A global maximum at an interior point requires checking both left and right derivatives or boundary behavior of piecewise defined functions.
<p>This question requires analyzing when \(x = 1\) is a global maximum. The domain constraint \(b^2 - 2 > 0\) requires \(|b| > \sqrt{2}\).</p><p>For \(x \leq 1\): analyze \(f'(x) = \frac{3}{2}x^2 - 2x + 10\)</p><p>For \(x > 1\): the function \(4 - 2^x + \log_2(b^2 - 2)\) is decreasing</p><p>For \(x = 1\) to be a maximum, we need \(f(1) \geq \lim_{x \to 1^+} f(x)\)</p><p>This gives the constraint on \(b\): \(b \in [-\sqrt{30}, -2) \cup (2, \sqrt{30}]\)</p>
Correct Answer: D