`f(x) = sec(x - pi/2), (0, 4pi)` Find the points of inflection.

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Chapter 3, 3.4 - Problem 27 - Calculus of a Single Variable (10th Edition, Ron Larson).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to evaluate the points of inflection of the function, hence, you need to solve for x the equation f''(x) = 0.

First, you need to evaluate the first derivative, using chain rule, such that:

`f'(x) = sec(x-pi/2)tan(x-pi/2)`

You need to evaluate the second derivative, using product rule:

`f''(x) = sec'(x-pi/2)tan(x-pi/2) + sec(x-pi/2)tan'(x-pi/2)`

`f''(x) = sec(x-pi/2)tan^2(x-pi/2) + sec(x-pi/2)(1 +tan^2(x-pi/2))`

Factoring out  `sec(x-pi/2)` yields:

`f''(x) =  sec(x-pi/2)(2tan^2(x-pi/2) + 1)`

You need to solve for x equation f''(x) = 0.

`sec(x-pi/2) = 0 => 1/(cos(x - pi/2)) = 0` impossible for `x in (0,4pi)`

`(2tan^2(x-pi/2) + 1) = 0 => (2tan^2(x-pi/2) =- 1` impossible for `x in (0,4pi)`

Hence, evaluating the inflection points in `(0,4pi),` yields that the function has no inflection points.

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