f(x)=1+x0.6(1+x)0.6
f′(x)=(1+x0.6)20.6(1+x)−0.4(1+x0.6)−0.6x0.4(1+x)0.6
=0.6(1+x)−0.4(1+x0.6)2[1+x0.6−x−0.4(1+x)2]=(1+x)0.4(1+x0.6)20.6(x0.4−1)
Clearly it is always negative in [0,1]
minimum value would be at x=1.
1+(1)0.6(1+1)0.6=220.6=2−0.4=k
& maximum values would be at x=0
1+(0)0.6(1+0)0.6=11=K
Hence (2−0.4,1)