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**

Graph confirms the concavity.

**(Appreciate if found helpful)**

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