y=x+cos^2 [0;pi/2] mn giúp em với

$$\eqalign{ & y = x + {\cos ^2}x\,\,\,\,x \in \left[ {0;{\pi \over 2}} \right] \cr & y' = 1 - 2\sin x\cos x = 0 \cr & \Leftrightarrow \sin 2x = 1 \cr & \Leftrightarrow 2x = {\pi \over 2} + k2\pi \Leftrightarrow x = {\pi \over 4} + k\pi \,\,\left( {k \in Z} \right) \cr & x \in \left[ {0;{\pi \over 2}} \right] \Rightarrow x = {\pi \over 4} \cr & y\left( 0 \right) = 0 + {\cos ^2}0 = 1 \cr & y\left( {{\pi \over 2}} \right) = {\pi \over 2} + {\cos ^2}{\pi \over 2} = {\pi \over 2} \cr & y\left( {{\pi \over 4}} \right) = {\pi \over 4} + {\cos ^2}{\pi \over 4} = {\pi \over 4} + {1 \over 2} \cr & \Rightarrow \mathop {\min }\limits_{\left[ {0;{\pi \over 2}} \right]} y = 1;\,\,\mathop {\max }\limits_{\left[ {0;{\pi \over 2}} \right]} y = {\pi \over 2} \cr}$$