Applications of Derivatives
Concavity and Inflection Points
Grade 12
Question:
<p>If the graph of the function \(f(x) = 3x^4 + 2x^3 + ax^2 - x + 2\) is concave up for all real values of \(x\), then the value of \(a\) is</p>
<p>(a) \(a > \frac{1}{2}\)</p>
<p>(b) \(a > 0\)</p>
<p>(c) \(a < -2\)</p>
<p>(d) \(a < \frac{1}{3}\)</p>
Step-by-Step Solution
Key Concept: A function is concave up when its second derivative is positive for all values in the domain. For a quadratic to be always positive, its discriminant must be negative.
<p><strong>Step 1:</strong> Given $f(x) = 3x^4 + 2x^3 + ax^2 - x + 2$</p><p><strong>Step 2:</strong> Calculate the first derivative: $f'(x) = 12x^3 + 6x^2 + 2ax - 1$</p><p><strong>Step 3:</strong> Calculate the second derivative: $f''(x) = 36x^2 + 12x + 2a$</p><p><strong>Step 4:</strong> For the graph to be concave up for all real $x$, we need $f''(x) > 0$ for all $x \in \mathbb{R}$</p><p><strong>Step 5:</strong> This means $36x^2 + 12x + 2a > 0$ for all $x$</p><p><strong>Step 6:</strong> For a quadratic $Ax^2 + Bx + C > 0$ for all $x$, we need $A > 0$ and discriminant $\Delta < 0$</p><p><strong>Step 7:</strong> Here $A = 36 > 0$ ✓, and $\Delta = 12^2 - 4(36)(2a) = 144 - 288a < 0$</p><p><strong>Step 8:</strong> From $144 - 288a < 0$, we get $144 < 288a$, so $a > \frac{144}{288} = \frac{1}{2}$</p><p>∴ Answer is (a).</p>
Correct Answer: A