The function
f(x) = ${{4{x^3} - 3{x^2}} \over 6} - 2\sin x + \left( {2x - 1} \right)\cos x$ :

Given, $f(x) = {{4{x^3} - 3{x^2}} \over 6} - 2\sin x + (2x - 1)\cos x$<br/><br/>$f'(x) = {{12{x^2} - 6x} \over 6} - 2\cos x + (2x - 1)( - \sin x) + \cos x(2)$<br/><br/>$= (2{x^2} - x) - 2\cos x - 2x\sin x + \sin x + 2\cos x$<br/><br/>$= 2{x^2} - x - 2x\sin x + \sin x$<br/><br/>$= 2x(x - \sin x) - 1(x - \sin x)$<br/><br/>$f'(x) = (2x - 1)(x - \sin x)$<br/><br/>for $x > 0,x - \sin x > 0$<br/><br/>$x < 0,x - \sin x < 0$<br/><br/>for $x \in ( - \infty ,0] \cup \left[ {{1 \over 2},\infty } \right),f'(x) \ge 0$<br/><br/>for $x \in \left[ {0,{1 \over 2}} \right],f'(x) \le 0$<br/><br/>Hence, f(x) increases in $\left[ {{1 \over 2},\infty } \right)$.