<p>Consider  the function \(f\) with the second derivative \(f''(x) = 3x  - 1\). The graph of \(f\) has a minimum point at \(\mathrm{A}(2,4)\) and a maximum point at \(\mathrm{B} \left( - \frac{4}{3}, \frac{358}{27} \right)\).</p>
<ol class="abc-list"><li><div class="list-item-paragraph">Use the second derivative to justify that&nbsp;\(\mathrm{B}\) is a maximum.<span class="inline-marks">[3]</span></div></li><li><div class="list-item-paragraph">Given that&nbsp;\(f' = \frac{3}{2}x^2 - x + p\), show that&nbsp;\(p = -4\).<span class="inline-marks">[4]</span></div></li><li><div class="list-item-paragraph">Find \(f(x)\).<span class="inline-marks">[7]</span></div></li></ol>

Consider  the function \(f\) with the second derivative \(f''(x) = 3x  - 1\). The graph of \(f\) has a minimum point at \(\mathrm{A}(2,4)\) and a maximum point at \(\mathrm{B} \left( - \frac{4}{3}, \frac{358}{27} \right)\).

Use the second derivative to justify that \(\mathrm{B}\) is a maximum.

Given that \(f' = \frac{3}{2}x^2 - x + p\), show that \(p = -4\).

Applications of Differential Calculus

International Baccalaureate Standard Level Mathematics