Limits, Continuity & Differentiability
Lagrange's Mean Value Theorem Applications
Grade 12
Question:
<p>Let <span class="math">f(x)</span> satisfy the requirements of Lagrange's mean value theorem in <span class="math">[0, 2]</span>. If <span class="math">f(0) = 0</span> and <span class="math">|f'(x)| \leq \frac{1}{2}</span> for all <span class="math">x \in [0, 2]</span>, then</p>
<p>(a) <span class="math">f(x) < 2</span></p>
<p>(b) <span class="math">|f(x)| < 2x - 4</span></p>
<p>(c) <span class="math">|f(x)| < 1</span></p>
<p>(d) <span class="math">f(x) = 3</span> for at least one <span class="math">x \in [0, 2]</span></p>
Step-by-Step Solution
Key Concept: Apply Lagrange's Mean Value Theorem on the interval [0, x] and use the bound on the derivative to bound the function value.
<p><strong>Solution:</strong> Let <span class="math">x \in (0, 2)</span>. Since <span class="math">f(x)</span> satisfies the requirements of Lagrange's mean value theorem in <span class="math">[0, 2]</span>, it also satisfies in <span class="math">[0, x]</span>. Consequently, there exists <span class="math">c \in (0, x)</span> such that <span class="math">f'(c) = \frac{f(x) - f(0)}{x - 0} = \frac{f(x)}{x}</span>. Therefore, <span class="math">|f(x)| = |f'(c)| \cdot x \leq \frac{1}{2} \cdot x</span>. Since <span class="math">x \in (0, 2)</span>, we have <span class="math">\frac{x}{2} \leq 1</span>, so <span class="math">|f(x)| \leq 1</span>.</p>
Correct Answer: C