Find the points on the curve y = 2x³ + 3x² -12x + 8 where the tangent is horizontal?

Solution:

Given

y = 2x³ + 3x² - 12x + 8 --->(1)

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

dy/dx = 6x² + 6x - 12 ---> (2)

Equation (2) is ‘slope if 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) + 8

y = -16 + 12 + 24 + 8

y = 28

When x = 1,

y = 2(1)³ + 3(1)² -12(1) + 8

y = 2 + 3 - 12 + 8

y = 1

Therefore, required points are (-2, 28) and (1, 1).

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Find the points on the curve y = 2x³ + 3x² - 12x + 8 where the tangent is horizontal?

Summary:

The points on the curve y = 2x³ + 3x² - 12x + 8, where the tangent is horizontal are (-2, 28) and (1, 1).