`f(x) = sin(x) + cos(x) [0, 2pi]` Find the points of inflection and discuss the concavity of the graph of the function.

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Chapter 3, 3.4 - Problem 28 - Calculus of a Single Variable (10th Edition, Ron Larson).
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Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

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This function is fully differentiable on a given interval. We can use the second derivative to determine concavity and inflection points. Let's find f'' and where its positive, negative or zero:

`f''(x) = -sin(x) - cos(x) = -sqrt(2)*cos(x-pi/4).`

This is zero at `x-pi/4 = pi/2+k*pi,` or `x = (3*pi)/4 + k*pi` for any integer k. There are two such x at [0, 2`pi` ], `x=(3pi)/4` and `x=(7pi)/4.`

f''(x) is negative on `(0, (3pi)/4),` positive on `((3pi)/4, (7pi)/4)` and is positive again on `((7pi)/4, 2pi).` Therefore f(x) is concave downward on `(0, (3pi)/4)` and on `((7pi)/4, 2pi),` and is concave upward on `((3pi)/4, (7pi)/4).`

The points of inflection are `(3pi)/4` and `(7pi)/4.`

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