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....

Printable View

- July 27th 2008, 08:29 PMzintoreinflection 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.... - July 27th 2008, 08:37 PMmr fantastic
- July 27th 2008, 08:38 PMzintoresorrry 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??

- July 27th 2008, 08:49 PMTKHunny
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. - July 28th 2008, 07:47 AMsquarerootof2
i thought by definition inflection points are where f''(x)=0. is that not true?