At which points on the curve y = x^3 - 3x^2 - | vedclass

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Question

(A) Given the curve equation: $y = x^3 - 3x^2 - 9x + 7$. Differentiating with respect to $x$: $\frac{dy}{dx} = 3x^2 - 6x - 9$. The tangent is parallel

Vedclass · Accepted Answer

$(-1, 12), (3, -20)$

Vedclass · Answer

(A) Given the curve equation: $y = x^3 - 3x^2 - 9x + 7$. Differentiating with respect to $x$: $\frac{dy}{dx} = 3x^2 - 6x - 9$. The tangent is parallel to the $x$-axis when its slope is zero,i.e.,$\frac{dy}{dx} = 0$. Setting the derivative to zero: $3x^2 - 6x - 9 = 0$. Dividing by $3$: $x^2 - 2x - 3 = 0$. Factoring the quadratic equation: $(x - 3)(x + 1) = 0$. This gives $x = 3$ and $x = -1$. For $x = -1$: $y = (-1)^3 - 3(-1)^2 - 9(-1) + 7 = -1 - 3 + 9 + 7 = 12$. For $x = 3$: $y = (3)^3 - 3(3)^2 - 9(3) + 7 = 27 - 27 - 27 + 7 = -20$. Thus,the points are $(-1, 12)$ and $(3, -20)$.

At which points on the curve $y = x^3 - 3x^2 - 9x + 7$ is the tangent parallel to the $x$-axis?

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