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Identify the inflection point for the function f(x)=8sinx+2x^2
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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.
Posted by giorgiana1976 on July 22, 2011 at 1:21 AM (Answer #1)
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