Find points on the curve y = 2x³ + 3x² - 12x + 4 where the tangent is horizontal?

Solution:

Given

y = 2x³ + 3x² -12x + 4 --- (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

6(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) + 4

y = -16 + 12 + 24 + 4

y = 24

When x = 1,

y = 2(1)³ + 3(1)² - 12(1) + 4

y = 2 + 3 - 12 + 4

y = -3

Therefore, the required points are (-2, 24) and (1, -3).

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Find points on the curve y = 2x³ + 3x² - 12x + 4 where the tangent is horizontal?

Summary:

The points on the curve y = 2x³ + 3x² - 12x + 1, where the tangent is horizontal are (-2, 24) and (1, -3).