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".
