Given, function is y=[x(x−2)]2=[x2−2x]2 On differentiating both sides w.r.t. x, we get
dy
dx
=2(x2−2x)
d
dx
(x2−2x) =2(x2−2x)(2x−2)=4x(x−2)(x−1) On putting
dy
dx
=0, we get 4x(x−2)(x−1)=0⇒x=0,1 and 2 Now, we find interval in which f(x) is strictly increasing or strictly decreasing.
Interval
dy
dx
=4x(x−2)(x−1)
Sign of f′(x)
(−∞,0)
(−)(−)(−)
−ve
(0,1)
(+)(−)(−)
+ve
(1,2)
(+)(−)(+)
−ve
(2,∞)
(+)(+)(+)
+ve
Hence, y is strictly increasing in (0,1) and (2,∞). Also, y is a polynomial function, so it continuous at X=0,1 and 2 . Hence, y is increasing in [0,1]∪[2,∞].