Limits, Continuity & Differentiability
Differentiability of Absolute Value Functions
Grade 12
Question:
<p>If <span class="math">f(x) = |x^2 - 3|x| + 2|</span>, then which of the following is/are true?</p>
<p>(a) <span class="math">f'(x) = 2x - 3</span> for <span class="math">x \in (0, 1) \cup (2, \infty)</span></p>
<p>(b) <span class="math">f'(x) = 2x + 3</span> for <span class="math">x \in (-\infty, -2) \cup (-1, 0)</span></p>
<p>(c) <span class="math">f'(x) = -2x - 3</span> for <span class="math">x \in (-2, -1)</span></p>
<p>(d) None of the above</p>
Step-by-Step Solution
Key Concept: Handle nested absolute values by analyzing the sign of inner expressions on different intervals.
<p><strong>Step 1:</strong> Analyze the nested absolute values. For <span class="math">x > 0</span>:</p><p>$$f(x) = |x^2 - 3x + 2| = |(x-1)(x-2)|$$</p><p><strong>Step 2:</strong> Determine sign on different intervals:</p><p>For <span class="math">x \in (0, 1) \cup (2, \infty)</span>: <span class="math">x^2 - 3x + 2 > 0</span>, so <span class="math">f(x) = x^2 - 3x + 2</span> and <span class="math">f'(x) = 2x - 3</span>. ✓</p><p><strong>Step 3:</strong> For <span class="math">x < 0</span>:</p><p>$$f(x) = |x^2 + 3x + 2| = |(x+1)(x+2)|$$</p><p>For <span class="math">x \in (-\infty, -2) \cup (-1, 0)</span>: <span class="math">x^2 + 3x + 2 > 0</span>, so <span class="math">f'(x) = 2x + 3</span>. ✓</p><p>For <span class="math">x \in (-2, -1)</span>: <span class="math">x^2 + 3x + 2 < 0</span>, so <span class="math">f(x) = -x^2 - 3x - 2</span> and <span class="math">f'(x) = -2x - 3</span>. ✓</p>
Correct Answer: A, B, C