Question

Determine the open intervals on which the graph is concave upward or concave downward. (Enter your answers using interval notation. If an answer does not exist, enter DNE.)

y = 3x +

2

sin x

,    (−π, π)

Answer 1

Solution:

Given

Differentiating with respect to x

Again differentiating with respect to x

Now find where y’’=0 or undefined in the interval [-π,π]

It is 0 in three places and there is no point in the interval at which it is undefined.

Now test the intervals determined by y’’=0 and find concavity based on the conditions if

y’’>0 means concave up

y’’<0 means concave down

Interval (-π,0):

Interval (0,π):

Hence for interval (-π,0) given function concaves up and for interval (0,π) it concaves down.

Graph confirms the concavity.

(Appreciate if found helpful)