Find the values of x for which f(x) = [x(x - 2)]2 is an increasing function. Also, find the points on the curve where the tangent is parallel to x-axis.

$$\begin{gathered} \mathrm{f} ( \mathrm{x} ) = {\left( \mathrm{x} \left( \mathrm{x} - 2 \right)\right)}^2 = {\mathrm{x}}^2 \left( {\mathrm{x}}^2 - 4 \mathrm{x} + 4 \right) = {\mathrm{x}}^4 - 4 {\mathrm{x}}^3 + 4 {\mathrm{x}}^2 \\ \mathrm{f} ( \mathrm{x} ) = 4 {\mathrm{x}}^3 - 12 {\mathrm{x}}^2 + 8 \mathrm{x} \\ \mathrm{f} ( \mathrm{x} ) = 4 \mathrm{x} \left( {\mathrm{x}}^2 - 3 \mathrm{x} + 2 \right) \\ = 4 \mathrm{x} \left( \mathrm{x} - 2 \right) \left( \mathrm{x} - 1 \right) \\ \mathrm{f} ( \mathrm{x} ) = 0 \Rightarrow \mathrm{x} = 0 \mathrm{or} 1, 2 \end{gathered}$$

So, the tangents to curve f ( x ) is parallel to the x-axis if  x = 0,  x = 1  or  x = 2.

Now points  0, 1  and  2 will divide the number line into 4 disjoint intervals 

$\left( -\infty , 0 \right), \left( 0, 1 \right), \left(1, 2 \right), \left(2, \infty \right)$

$$\begin{gathered} \mathrm{Now} \mathrm{in} \mathrm{the} \mathrm{intervals} \left( -\infty , 0 \right) \mathrm{and} \left( 1, 2 \right) \mathrm{f} ( \mathrm{x} ) < 0. \mathrm{So} \mathrm{thefunction} \mathrm{f} ( \mathrm{x} ) \\ \mathrm{is} \mathrm{strictly} \mathrm{decreasing} \mathrm{in} \mathrm{these} \mathrm{intervals}. \\ \mathrm{f} ( \mathrm{x} ) > 0 \mathrm{in} \mathrm{interval} ( 0, 1 ) \mathrm{and} ( 2, \infty ) \\ \mathrm{So} \mathrm{the} \mathrm{function} \mathrm{f} ( \mathrm{x} ) \mathrm{is} \mathrm{strictly} \mathrm{increasing} \mathrm{in} \mathrm{intervals} ( 0, 1 ) \mathrm{and} ( 2, \infty ) \\ \mathrm{Tangent} \mathrm{is} \mathrm{parallel} \mathrm{to} \mathrm{x}-\mathrm{axis} \mathrm{if} \frac{\mathrm{dy}}{\mathrm{dx}} = 0 \end{gathered}$$

Which gives us  x = 0,  1,  2

Hence,  x = 0,  y = 0

x = 1,  y = 0

x = 2,  y = 0

Required points are ( 0, 0 ),  ( 1, 1 ),  ( 2, 0 ).