question_answer

Show that the function given by f(x) = sinx is neither strictly decreasing nor strictly increasing in the interval \[(0,\,\,\pi ).\]

Answer:

Given, f(x) = sin.x On differentiating both sides w.r.t. x, we get             \[f'(x)=\cos x\] We know that, cos x > 0 when \[x\in \left( 0,\,\,\frac{\pi }{2} \right)\] and cos x < 0 when \[x\in \left( \frac{\pi }{2},\,\,\pi  \right).\] Thus, f(x) is strictly increasing in \[\left( 0,\,\,\frac{\pi }{2} \right)\] and strictly decreasing in \[\left( \frac{\pi }{2},\,\,\pi  \right).\] Hence, f(x) is neither strictly increasing nor strictly decreasing.