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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.

f(x) = x4 - 6x2

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So, the points of inflection are at x = ± 1. 

f”(x) = 0 ⇒ x2 = 1 ⇒ x = ± 1 

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

When x ∈ (-∞, -1), say x = -2 

f”(x) = 12(-2)2 – 12 = 48 – 12 = 36 > 0 

⇒ f(x) is concave upward in (-∞, -1) when x ∈ (-1, 1) say x = 0. 

f”(x) = 0 – 12 = -12 < 0 

⇒ f(x) convex upward.

⇒ x = -1 is a point of inflection. Again when x ∈ (-1, 1), the curve is convex upward and when x ∈ (1, ∞) say x = 2. 

f”(x) = (1, ∞) say x = 2. 

f”(x) = 12 (4) – 12 > 0 

⇒ the curve is concave upward. 

⇒ x = 1 is a point of inflection. 

Thus x = ± 1 are the points of inflection. 

At x = 1, f(x) = 1 – 6 = -5 

At x = -1, f(x) = (-1) – 6(-1) = 1 – 6 = -5 

The curve concave upward in (-∞, -1) u (1, ∞) and convex upward in (-1, 1) and the points of inflection are (1, -5) and (-1, -5).

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