Question

Find intervals of concavity and points of inflection for the following functions:

f(x) = sin x + cos x, 0 < x < 2π

Answer 1

f'(x) = cos x – sin x

f”(x) = – sin x – cos x

f'(x) = 0

⇒ sin x + cos x = 0

Critical points x = `(3pi)/4, (7pi)/4`

The intervals are `(0, (pi)/4),((3pi)/4, (7pi)/4)` and `((7pi)/4, 2pi)`

In the interval `(0, (3pi)/4)`, f(x) < 0 ⇒ curve is concave down.

In the interval `((3pi)/4, (7pi)/4)`, f'(x) > 0 ⇒ curve is concave up.

In the interval `((7pi)/4, 2pi)`, f'(x) < 0 ⇒ curve is concave down.

The curve is concave upward in `((3pi)/4, (7pi)/4)` and concave downward in `(0, (3pi)/4)` and `((7pi)/4, 2pi)`

f'(x) changes its sign when passing through x = `(3pi)/4` and x = `(7pi)/4`

Now `"f"((3pi)/4) = sin  (3pi)/4 + cos  (3pi)/4`

= `sqrt(2)/2 - sqrt(2)/2`

= 0

`"f"((7pi)/4) = sin  (7pi)/4 + cos  (7pi)/4`

= `- sqrt(2)/2 + sqrt(2)/2`

= 0

∴ The point of inflection are `((3pi)/4, 0)` and `((7pi)/4, 0)`.