Question

Determine the absolute minimum value of
f(x) = cos² (x) + sin(x)
on [0, 2π].

Answer 1

`f_(min)=f((3\pi)/(2))=-1.`

Explanation:

1) f'[x]=cos[x]-2sin[x]cos[x];

2) f'[x]=0, ⇒ cos[x]-2sin[x]cos[x]=0;

`\left \ [ {{cos[x]=0} \atop {sin[x]=0.5}} \right. \ = > \ \left \ [ {{x=(\pi)/(2)+\pi*n } \atop {x=(-1)^(n+1)*(\pi)/(6) +\pi*n}} \right.`

`3) \ x=(\pi)/(2) +\pi*n \ -min, \ = > f_(min)|_{x=(\pi)/(2) ;(3\pi)/(2) }:\\\left \ [ {{f[(\pi)/(2)]=1} \atop {f[(3\pi)/(2)] =-1}} \right. = > \ f_(min)=f[(3\pi)/(2)]=-1.`

4) finally, f(min)=-1.