Find the critical numbers of the function. (Enter your answers as a comma-separated list.)

Question

h(x) = sin² x + cos x  ,  0 < x < 2𝜋

Answer 1

h'(x)=2sin(x)cos(x)-sin(x)

set to 0 for critical points

have

sin(x)(2cos(x)-1)=0

solved when sin(x)=0 which is x=-0 or x=π

solved also when (2cos(x)-1)=0

cos(x)=1/2 when x=π/3 and x=5π/3

what values does h(x) have at critical domain values

h(x)=sin²(x)+cos(x)

x=0, h=1

x=π/3, h=5/4

x=π, h=-1

x=5π/3 h=5/4

Answer 2

Take derivative of h(x)

h'(x)=2sin(x)cos(x) - sin(x)

Set derivative=0 and solve for x

2sin(x)cos(x) - sin(x) → 2sin(x)(cos(x) -1)

0=2sin(x)(cos(x) - 1)

2sin(x)=0, x=π/2

cos(x) - 1=0, cos(x)=1, x=0 or 2π

Since π/2 is within interval of 0<x<2π, h(x) has one critical point of π/2