The stationary points of the curve y = 8x^2 - | vedclass

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Question

(A) Given curve is $y = 8x^2 - x^4 - 4$. To find the stationary points,we differentiate the function with respect to $x$: $\frac{dy}{dx} = 16x - 4x^3$

Vedclass · Accepted Answer

$(0, -4), (2, 12), (-2, 12)$

Vedclass · Answer

(A) Given curve is $y = 8x^2 - x^4 - 4$. To find the stationary points,we differentiate the function with respect to $x$: $\frac{dy}{dx} = 16x - 4x^3$. Stationary points occur where $\frac{dy}{dx} = 0$: $16x - 4x^3 = 0$ $4x(4 - x^2) = 0$ This gives $x = 0$ or $x^2 = 4$,which implies $x = 0, 2, -2$. Now,we find the corresponding $y$-values: For $x = 0$: $y = 8(0)^2 - (0)^4 - 4 = -4$. For $x = 2$: $y = 8(2)^2 - (2)^4 - 4 = 8(4) - 16 - 4 = 32 - 20 = 12$. For $x = -2$: $y = 8(-2)^2 - (-2)^4 - 4 = 8(4) - 16 - 4 = 32 - 20 = 12$. Thus,the stationary points are $(0, -4), (2, 12), (-2, 12)$.

The stationary points of the curve $y = 8x^2 - x^4 - 4$ are . . . . . .

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