Find points on the curve y=2x³ + 3x² - 12x + 1 where the tangent is horizontal?

Solution:

Given

y = 2x³ + 3x² - 12x + 1 --- (1)

Differentiate equation (1) w.r.t., x

dy/dx = 6x² + 6x - 12 --- (2)

Equation (2) is the ‘slope of the tangent at any point’.

Since the tangent is horizontal, its slope is zero.

∴ 6x² + 6x - 12 = 0

x² + x - 2 = 0

(x + 2)(x - 1) = 0

∴ The roots are x = -2 and x = 1

When x = -2,

y = 2(-2)³ + 3(-2)² - 12(-2) + 1

y = -16 + 12 + 24 + 1

y = 21

When x = 1,

y = 2(1)³ + 3(1)² -12(1) +1

y = 2 + 3 - 12 + 1

y = -6

Therefore, required points are (-2, 21) and (1, -6).

---

Find points on the curve y=2x³ + 3x² - 12x + 1 where the tangent is horizontal?

Summary:

The points on the curve y = 2x³ + 3x² - 12x + 1, where the tangent is horizontal are (-2, 21) and (1, -6).