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Identify the inflection point for the function f(x)=8sinx+2x^2

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greynose | Student, Undergraduate | Honors

Posted July 22, 2011 at 1:16 AM via web

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Identify the inflection point for the function f(x)=8sinx+2x^2

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giorgiana1976 | College Teacher | Valedictorian

Posted July 22, 2011 at 1:21 AM (Answer #1)

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To find the inflection point, w'll have to determine the second derivative of the function.

We'll differentiate with respect to x:

f'(x) = 8cos x + 4x

f"(x) = -8sin x + 4

Now, we'll set f"(x) equal to 0:

f"(x) = 0 <=> -8sin x + 4 = 0

We'll divide by 4 both sides:

-2sin x + 1 = 0

-2sin x = -1

sin x = 1/2

x = (-1)^k*arcsin(1/2) + k`pi`

x = (-1)^k*(`pi` /6) + k`pi`

The inflection point of the function is at x = (-1)^k*(` ` /6) +kPi.

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