Given the function `h(x)=sin^2(x)+cos(x)` in the interval `0<x<2pi`
We have to find the critical numbers of the function.
First take the derivative of the function and equate it to zero.
We get,
`h'(x)=2sin(x)cos(x)-sin(x)=0`
`sin(x)(2cos(x)-1)=0`
`sin(x)=0` or `2cos(x)-1=0`
sin(x)=0 implies x= npi
i.e we get x= pi in the interval 0<x<2pi
Now,
2cos(x)-1=0 implies cos(x)=1/2
So x= pi/3 and 5pi/3 (in the interval 0<x<2pi)
Hence the critical points are `x=pi/3, pi and (5pi)/3`
See eNotes Ad-Free
Start your 48-hour free trial to get access to more than 30,000 additional guides and more than 350,000 Homework Help questions answered by our experts.
Already a member? Log in here.