`y = x + 2/sin(x) (-pi, pi)` Determine the open intervals on whcih the graph is concave upward or downward.

Textbook Question

Chapter 3, 3.4 - Problem 14 - Calculus of a Single Variable (10th Edition, Ron Larson).
See all solutions for this textbook.

1 Answer | Add Yours

shumbm's profile pic

Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

Posted on

This function is undefined at 0, i.e. we have to consider it on `(-pi, 0)` and `(0, pi)` separately.

To determine concavity, we have to find the second derivative and its sign.

`f'(x) = 1 - 2(cosx)/(sin^2x),`

`f''(x) = -2*(-sinx*sin^2x - cosx*2sinx*cosx)/(sin^4x) =`

`= 2(sin^2x+2cos^2x)/(sin^3x) = 2(1+cos^2x)/(sin^3x).`

This is has the same sign as sinx, namely "-" for x in `(-pi, 0)` and "+" in` (0, pi).`

So the answer is: y is concave downwards on `(-pi, 0)` and is concave upwards on `(0, pi).`

We’ve answered 318,915 questions. We can answer yours, too.

Ask a question