Applications of Derivatives
Maxima and Minima
Grade 12
Question:
<p>Let <em>f</em>(<em>x</em>) = <em>a</em><em>x</em><sup>3</sup> + <em>b</em><em>x</em><sup>2</sup> + 6, where the coefficient of <em>x</em> and <em>x</em><sup>0</sup> are 0 and 6 respectively, <em>f</em>‴(<em>x</em>) = 24, and an extreme of <em>f</em>′(<em>x</em>) occurs at <em>x</em> = −1/6. Find the maximum value of <em>f</em>(<em>x</em>) for <em>x</em> ∈ [0, 2].</p>
Step-by-Step Solution
Key Concept: Since f(x) = ax³ + bx² + 6 has no x term, f'(x) = 3ax² + 2bx. The condition that f'(x) has an extremum at x = -1/6 means f''(-1/6) = 0, which combined with f'''(x) = 24 gives us a = 4. Then use f''(-1/6) = 0 to find b = 2, and evaluate f at critical points and endpoints.
<p><strong>Step 1: Find coefficients a and b</strong></p><p>Given f(x) = ax³ + bx² + 6, so f'(x) = 3ax² + 2bx and f''(x) = 6ax + 2b</p><p>From f'''(x) = 24: Since f'''(x) = 6a, we get 6a = 24 → <strong>a = 4</strong></p><p><strong>Step 2: Use the extremum condition</strong></p><p>f'(x) has an extremum at x = -1/6 means f''(-1/6) = 0:</p><p>6(4)(-1/6) + 2b = 0</p><p>-4 + 2b = 0 → <strong>b = 2</strong></p><p><strong>Step 3: Identify the function</strong></p><p>f(x) = 4x³ + 2x² + 6</p><p>f'(x) = 12x² + 4x = 4x(3x + 1)</p><p><strong>Step 4: Find critical points in [0, 2]</strong></p><p>f'(x) = 0 → x = 0 or x = -1/3</p><p>Only x = 0 lies in [0, 2]</p><p><strong>Step 5: Evaluate at critical point and endpoints</strong></p><p>f(0) = 6</p><p>f(2) = 4(8) + 2(4) + 6 = 32 + 8 + 6 = <strong>46</strong></p><p>∴ Maximum value of f(x) on [0, 2] is <strong>46</strong></p>
Correct Answer: 46