Question

Find points at which the tangent to the curve y = x³ – 3x² – 9x + 7 is parallel to the x-axis.

Answer 1

\(\frac{dy}{dx}\) = 3x² – 6x – 9, slope of the tangent 

since the tangent is parallel to x-axis \(\frac{dy}{dx}\)= 0 

3x² – 6x – 9 = 0 

3(x + 1) (x – 3) = 0, x = 3, x = -1 

when x = 3, y = 27 – 27 – 27 - 7 = -20 

when x = -1, y = -1 - 3 + 9 + 7 = 12 

The points at which the tangent parallel to x – axis are (3, -20) and (-1, 12).