If #f(x)=x^4(x-1)^3#, then the Product Rule says
#f'(x)=4x^3(x-1)^3+x^4*3(x-1)^2#
#=x^3*(x-1)^2*(4(x-1)+3x)#
#=x^3*(x-1)^2*(7x-4)#
Setting this equal to zero and solving for #x# implies that #f# has critical numbers (points) at #x=0,4/7,1#.
Using the Product Rule again gives:
#f''(x)=d/dx(x^3*(x-1)^2) * (7x-4)+x^3*(x-1)^2*7#
#=(3x^2*(x-1)^2+x^3*2(x-1)) * (7x-4) + 7x^3 * (x-1)^2#
#=x^2 * (x-1) * ((3x-3+2x) * (7x-4) + 7x^2-7x)#
#=x^2 * (x-1) * (42x^2-48x+12)#
#=6x^2 * (x-1) * (7x^2-8x+2)#
Now #f''(0)=0#, #f''(1)=0#, and #f''(4/7)=576/2401>0#.
The Second Derivative Test therefore implies that the critical number (point) #x=4/7# gives a local minimum for #f# while saying nothing about the nature of #f# at the critical numbers (points) #x=0,1#.
In actuality, the critical number (point) at #x=0# gives a local maximum for #f# (and the First Derivative Test is strong enough to imply this, even though the Second Derivative Test gave no information) and the critical number (point) at #x=1# gives neither a local max nor min for #f#, but a (one-dimensional) "saddle point".
