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

1 Answer | Add Yours

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.

We’ve answered 333,647 questions. We can answer yours, too.

Ask a question