The abscissa of the points,where the tangent | vedclass

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(C) Given the curve equation is $y=x^3-3x^2-9x+5$. To find the slope of the tangent,we differentiate $y$ with respect to $x$: $\frac{dy}{dx} = 3x^2-6x

Vedclass · Accepted Answer

$x=-1$ and $3$

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(C) Given the curve equation is $y=x^3-3x^2-9x+5$. To find the slope of the tangent,we differentiate $y$ with respect to $x$: $\frac{dy}{dx} = 3x^2-6x-9$. Since the tangent is parallel to the $X$-axis,its slope must be zero: $\frac{dy}{dx} = 0 \Rightarrow 3x^2-6x-9 = 0$. Dividing by $3$,we get: $x^2-2x-3 = 0$. Factoring the quadratic equation: $(x-3)(x+1) = 0$. Thus,the values of $x$ are $x=3$ and $x=-1$.

The abscissa of the points,where the tangent to the curve $y=x^3-3x^2-9x+5$ is parallel to the $X$-axis,are

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