Question

Show that f(x) = cos x is a decreasing function on (0, π), increasing in (−π, 0) and neither increasing nor decreasing in (−π, π).

Answer 1

$$\begin{gathered} \mathrm{Here}, \\ f\left(x\right)=\cos x \\ \text{Domain of cos}\mathit{\text{x}}\text{is}\left(-\mathrm{\pi },\mathrm{\pi }\right)\text{.} \\ \Rightarrow f\text{'}\left(x\right)=-\sin x \\ \mathrm{For}x\in \left(-\mathrm{\pi },0\right),\sin x<0\left[\because \mathrm{sine}\mathrm{function}\mathrm{is}\mathrm{negative}\mathrm{in}\mathrm{third}\mathrm{and}\mathrm{fourth}\mathrm{quadrant}\right] \\ \Rightarrow -\sin x>0 \\ \Rightarrow f\text{'}\left(x\right)>0 \\ \mathrm{So},\cos \mathrm{x}\mathrm{is}\mathrm{increasing}\mathrm{in}\left(-\mathrm{\pi },0\right). \\ \mathrm{For}\mathit{}x\in \left(0,\mathrm{\pi }\right)\text{),}\sin \mathrm{x}>0\left[\because \mathrm{sine}\mathrm{function}\mathrm{is}\mathrm{positive}\mathrm{in}\mathrm{first}\mathrm{and}\mathrm{second}\mathrm{quadrant}\right] \\ \Rightarrow -\sin x<0 \\ \Rightarrow f\text{'}\left(x\right)<0 \\ \text{So,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is decreasing on}\left(0,\mathrm{\pi }\right)\text{.} \\ \text{Thus,}\mathit{\text{f}}\text{(}\mathit{\text{x}})\text{is neither increasing nor decreasing in}\left(-\mathrm{\pi },\mathrm{\pi }\right)\text{.} \end{gathered}$$