Limits, Continuity & Differentiability
Lagrange's Mean Value Theorem
Grade 12
Question:
<p>Suppose that <i>f</i>(0) = −3 and <i>f</i>′(<i>x</i>) ≤ 5 for all values of <i>x</i>. Then, the largest value which <i>f</i>(2) can assume is ________.</p>
Step-by-Step Solution
Key Concept: Apply Lagrange's Mean Value Theorem to relate the difference in function values to the derivative bound to find the maximum possible value of f(2).
<p><strong>Step 1:</strong> Using Lagrange's Mean Value Theorem on the interval [0, 2]:</p><p>$$\frac{f(2) - f(0)}{2 - 0} = f'(c) \text{ for some } c \in (0, 2)$$</p><p><strong>Step 2:</strong> Given that <i>f</i>(0) = −3 and <i>f</i>′(<i>x</i>) ≤ 5 for all <i>x</i>:</p><p>$$f'(c) \leq 5$$</p><p><strong>Step 3:</strong> Substituting into the LMVT equation:</p><p>$$\frac{f(2) - (-3)}{2} \leq 5$$</p><p>$$\frac{f(2) + 3}{2} \leq 5$$</p><p>$$f(2) + 3 \leq 10$$</p><p>$$f(2) \leq 7$$</p><p><strong>Step 4:</strong> The largest value that <i>f</i>(2) can assume is 7.</p><p>∴ Answer is <strong>7</strong>.</p>
Correct Answer: 7