1. ## inflection points

f(x)=x+sinx [0,2pi]
f'(x)=1 +cosx
f''(x)= -sinx
0=-sinx
x= 0, pi and 2pi
i dont know if i can call 0 and 2pi inflection points since they are end points of the closed interval....

2. Originally Posted by zintore
f(x)=x+sinx [0,2pi]
f'(x)=1 +cosx
f''(x)= -sinx
0=-sinx
x= 0, pi and 2pi
i dont know if i can call 0 and 2pi inflection points since they are end points of the closed interval....
Critical points are found by solving f'(x) = 0, NOT f''(x) = 0.

Note also, that a point has coordinates, that is, an x-coordinate AND a y-coordinate.

3. ## sorrry still not answered

but i was asking for the inflection points, where the graph changes concavity. can i say it changes concavity at an end point??

4. 1) The derivative does not exist at an endpoint.

2) Nothing can "change" at an endpoint, since there is only one side.

3) Extrema can occur at endpoints, but the calculus will find them only by design or coincidence. You must check the endpoints yourself.

5. i thought by definition inflection points are where f''(x)=0. is that not true?