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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) = 2x3 + 5x2 – 4x

1 Answer

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by (49.4k points)
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Best answer

y = f (x) = 2x3 + 5x2 – 4x 

f'(x) = 6x2 + 10x – 4 

f”(x) = 12x + 10; f'” (x) = 12 

f”(x) = 0 ⇒ 12x + 10 = 0

Consider x in (-∞, -5/6) say x = -1 

f”(x)= -12 + 10 = -2 < 0 ⇒ the curve convex upward in the interval (-∞, -5/6) Consider x in (-5/6, ∞) say x = 0 f”(x) = 0 + 10 = 10 > 0 

⇒ the curve is concave upward in (-5/6, -∞) 

Thus, the curve is concave upward in (-5/6, ∞) and convex upward in (-∞, – 5/6)

The point of inflection is ((-5/6), (305/54))

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