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(A) The equation of the given curve is $y=x^{3}-3x^{2}-9x+7$. The slope of the tangent is given by the derivative $\frac{dy}{dx} = 3x^{2}-6x-9$. The t

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The equation of the given curve is $y=x^{3}-3x^{2}-9x+7$. The slope of the tangent is given by the derivative $\frac{dy}{dx} = 3x^{2}-6x-9$. The tangent is parallel to the $x$-axis if the slope is zero,i.e.,$\frac{dy}{dx} = 0$. $3x^{2}-6x-9 = 0 \Rightarrow x^{2}-2x-3 = 0$. $(x-3)(x+1) = 0$,which gives $x=3$ or $x=-1$. For $x=3$,$y = (3)^{3}-3(3)^{2}-9(3)+7 = 27-27-27+7 = -20$. For $x=-1$,$y = (-1)^{3}-3(-1)^{2}-9(-1)+7 = -1-3+9+7 = 12$. Thus,the required points are $(3, -20)$ and $(-1, 12)$.

Find the points at which the tangent to the curve $y=x^{3}-3x^{2}-9x+7$ is parallel to the $x$-axis.

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