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

7) Find points at which the tangent to the curvey = x^3 - 3 x^2 - 9x +7 is parallel to the x-axis.

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We are given  :                   

                                                      y = x^3 - 3 x^2 - 9x +7

Differentiating the equation with respect to x,  we get :

                                                     \frac{dy}{dx}\ =\ 3x^2\ -\ 6x\ -\ 9\ +\ 0

or                                                           =\ 3\left ( x^2\ -\ 2x\ -\ 3 \right )

or                                                \frac{dy}{dx}\ =\ 3\left ( x+1 \right )\left ( x-3 \right )

It is given that tangent is parallel to the x-axis, so the slope of the tangent is equal to 0.

 So, 

                                                        \frac{dy}{dx}\ =\ 0

or                                                    0\ =\ 3\left ( x+1 \right )\left ( x-3 \right )

Thus,                                       Either       x =  -1     or          x = 3

When x = -1 we get y = 12          and         if  x =3 we get y = -20

So the required points are   (-1, 12)  and  (3, -20).

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