Question

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

Answer 1

The equation of the curve is $y=x^3-3x^2-9x+7$ ....(1)

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

Now, the tangent is parallel to X-axis, then slope of the tangent is zero or we can say $\frac{dy}{dx}=0$

$\Rightarrow 3x^2-6x-9=0\Rightarrow 3(x^2-2x-3)=0$

$\Rightarrow (x-3)(x+1)=0\Rightarrow x=3,-1$

When x=3, then from Eq. (1), we get

$y=3^3-\left(3\right).(3{)}^2-9,3+7=27-27-27+7=-20$

When x=-1, then from Eq. (i), we get

$y=(-1{)}^3.-3(-1{)}^2-9,(-1)+7=-1-3+9+7=12$

Hence, the points at whihc the tangetn is parallel to X-axis are (3,-20) and (-1,12)