Show that the function `f(x) = cos x`
(i) is strictly decreasing function in `]0,pi[`.
(ii) is neither increasing nor decreasing in `]0,2 pi[`.
(iii) is neither increasing nor decreasing in `]0,2 pi[`.

Explanation

`f(x)=cosx`
`f ′(x)=−sinx`
`(i) x in(0,pi)
⇒sinx gt 0⇒−sinx lt 0`
=`f ′(x) lt 0`so f(x) is strictly decreasing on `(0,pi)`

(ii) `x in(pi,2pi)⇒sinx<0
⇒−sinx gt 0`
⇒`f ′(x) gt 0`
so f(x) is strictly increasing on `(pi,2pi)`
(iii) As `f ′(x) lt 0 for x∈(0,pi)`
and `f ′(x) gt 0 for x∈(0,2pi)`
Hence f(x) is neither increasing nor decreasing on `(0,2pi)`.