If x=-2 and x=4 are the extreme points of y=x | vedclass

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(A) Given the function $y=x^3-\alpha x^2-\beta x+5$. Find the derivative with respect to $x$: $\frac{dy}{dx} = 3x^2 - 2\alpha x - \beta$. Since $x=-2$

Vedclass · Accepted Answer

$\alpha=3, \beta=24$

Vedclass · Answer

(A) Given the function $y=x^3-\alpha x^2-\beta x+5$. Find the derivative with respect to $x$: $\frac{dy}{dx} = 3x^2 - 2\alpha x - \beta$. Since $x=-2$ and $x=4$ are extreme points,the derivative must be zero at these values. For $x=-2$: $3(-2)^2 - 2\alpha(-2) - \beta = 0 \implies 12 + 4\alpha - \beta = 0 \implies 4\alpha - \beta = -12$ (Equation $1$). For $x=4$: $3(4)^2 - 2\alpha(4) - \beta = 0 \implies 48 - 8\alpha - \beta = 0 \implies 8\alpha + \beta = 48$ (Equation $2$). Adding Equation $1$ and Equation $2$: $(4\alpha - \beta) + (8\alpha + \beta) = -12 + 48 \implies 12\alpha = 36 \implies \alpha = 3$. Substitute $\alpha = 3$ into Equation $1$: $4(3) - \beta = -12 \implies 12 - \beta = -12 \implies \beta = 24$. Thus,$\alpha=3$ and $\beta=24$.

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

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