Question

If $x = -2$ and $x = 4$ are the extreme points of the function $y = x^3 - \alpha x^2 - \beta x + 5$, then the values of $\alpha$ and $\beta$ are respectively

• A. $\alpha = 3$, $\beta = 24$
• B. $\alpha = -24$, $\beta = -3$
• C. $\alpha = -3$, $\beta = -24$
• D. $\alpha = 24$, $\beta = 3$

Hint:

Since $-2$ and $4$ are roots of the quadratic derivative equation $3x^2 - 2\alpha x - \beta = 0$, you can use Vieta's formulas! The sum of the roots is equal to $-(\frac{\text{coefficient of } x}{\text{coefficient of } x^2})$: $(-2) + 4 = \frac{2\alpha}{3} \implies 2 = \frac{2\alpha}{3} \implies \alpha = 3$. This gives you half of the answer immediately without setting up full equations!

Answer 1

The Correct Option is A

Step 1: Use the meaning of extreme points.
At an extreme point the first derivative is zero, and we are told these occur at $x = -2$ and $x = 4$ for $y = x^3 - \alpha x^2 - \beta x + 5$.
Step 2: Differentiate.
$\dfrac{dy}{dx} = 3x^2 - 2\alpha x - \beta$.
Step 3: Apply x = -2.
$3(4) + 4\alpha - \beta = 0 \Rightarrow 4\alpha - \beta = -12$. (Equation 1)
Step 4: Apply x = 4.
$3(16) - 8\alpha - \beta = 0 \Rightarrow 8\alpha + \beta = 48$. (Equation 2)
Step 5: Add the two equations.
Adding eliminates $\beta$: $12\alpha = 36$, so $\alpha = 3$.
Step 6: Back-substitute for beta.
From Equation 2, $8(3) + \beta = 48 \Rightarrow \beta = 24$. So $\alpha = 3,\ \beta = 24$, which is option 1 and matches the key.
\[ \boxed{\alpha = 3,\ \beta = 24} \]