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
...
See This Answer Now
Start your subscription to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.
Already a member? Log in here.
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`