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+1 vote
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in Applications of Differential Calculus by (48.7k points)
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Find the intervals of concavity and the points of inflection of the following function.

y = 12x2 – 2x3 – x4

1 Answer

+1 vote
by (49.4k points)
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Best answer

So the intervals are (-∞, -2), (-2, 1)(1, ∞) 

When x ∈ (-∞, -2) say x = -3

d2y/dx2 = 12(2 + 3 - 9) = 12(-4) > 0

The curve is concave upward. 

When x ∈ (-2, 1) say x = 0.

d2y/dx2 = 12(2 - 0 - 0) = 24 > 0

⇒ The curve is concave upward.

x = -2 is a point of inflection.

When x ∈ (1,∞) say x = 2

d2y/dx2 = 12(2 - 2 - 4) = 12(-4) < 0

⇒ The curve is convex upward. So, x = 1 is a point of inflection. 

At x = – 2, y = 12(4) – 2(-8) – (16) = 48 + 16 – 16 = 48 

At x = 1, y = 12(1) – 2(1) – 1 = 12 – 2 – 1 = 9 

So, the points of inflection are (1, 9) and (-2, 48)

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