For extrema, #f'(x)=0.#
#f'(x)=(x-2)*3(x-5)^2+(x-5)^3*1=(x-5)^2{3x-6+x-5]=(4x-11)(x-5)^2.#
#f'(x)=0 rArr x=5!in [1,4],# so no need for further cosideration & #x=11/4.#
#f'(x)=(4x-11)(x-5)^2, rArr f''(x)=(4x-11)*2(x-5)+(x-5)^2*4=2(x-5){4x-11+2x-10}=2(x-5)(6x-21).#
Now, #f''(11/4)=2(11/4-5)(33/2-21)=2(-9/4)(-9/2)>0,# showing that, #f(11/4)=(11/4-2)(11/4-5)^3=(3/2)(-9/4)^3=-2187/128,# is Local Minima.
To find Global Values, we need #f(1)=(1-2)(1-5)^3=64,# & #f(4)=(4-2)(4-5)^3=-2.#
Hence, Global Minima #=min# { local minima, #f(1), f(4)}=min{-2187/128,64, -2}=min{-17.09, 64, -2}=-2187/128~=-17.09#
Global Maxima #=max# { local maxima ( which doesn't exist), #f(1), f(4)}=max{64, -2}=64.#
