Limits, Continuity & Differentiability
Differentiability and limits
Grade 12
Question:
<p>If \(g(x) = (x^2 + 2x + 3)f(x)\), \(f(0) = 5\) and \(\displaystyle\lim_{x \to 0}\left(\dfrac{f(x)-5}{x}\right) = 4\), then \(g'(0)\) is equal to:</p>
<p>(a) 22</p>
<p>(b) 18</p>
<p>(c) 23</p>
<p>(d) 25</p>
Step-by-Step Solution
Key Concept: Recognize that the given limit is the definition of f'(0) = 4, then apply the product rule: g'(x) = (x² + 2x + 3)f'(x) + f(x)·(2x + 2), evaluating at x = 0.
<p><strong>Step 1:</strong> Identify the given information.</p><p>Given: g(x) = (x² + 2x + 3)f(x), f(0) = 5, and lim[x→0] (f(x) - 5)/x = 4</p><p>The limit condition means f'(0) = 4 (by definition of derivative)</p><p><strong>Step 2:</strong> Apply the product rule to find g'(x).</p><p>g'(x) = d/dx[(x² + 2x + 3)] · f(x) + (x² + 2x + 3) · f'(x)</p><p>g'(x) = (2x + 2)f(x) + (x² + 2x + 3)f'(x)</p><p><strong>Step 3:</strong> Evaluate g'(0).</p><p>g'(0) = (2(0) + 2)f(0) + (0² + 2(0) + 3)f'(0)</p><p>g'(0) = 2 · 5 + 3 · 4</p><p>g'(0) = 10 + 12</p><p>∴ Answer: g'(0) = <strong>22</strong></p>
Correct Answer: A