f(x) = x5/5 – 5x5/3 + 4x for the critical points you need the first two derivatives, but first we simplify.
f(x)=x5(1/5-5/3)+4x=(3/15-25/15)x5+4x=(-22/15)x5+4x
f'(x)=-110x4/15+4 f'(x)=0 for 110x4/15=4 or x4=60/110 = 6/11for which the real roots are
x=±(6/11)1/4
To determine whether these are a min or a max we need f''(x) at this point.
f''(x)=-440x3/15, which equals zero only at x=0, the point of inflection, now
f''((6/11)1/4)=-440(6/11)1/4/15<0 and so the critical point x= (6/11)1/4is a maximum
f''(-(6/11)1/4)=440(6/11)1/4/15>0 and so the critical point x=-(6/11)1/4is a minimum
f(x) = x4/4 - x3 + 1
f'(x)=x3-3x2=x2(x-3)
f"(x)=3x2-6x
f'(x)=0 at x=0, and x=3
f"(0)=0 and f"(3)=9
This means that x=3 is a minimum and x=0 is a point of inflection
The curve looks like -x3 going from left to right through x=0 goes down to it min at x=3. Away from x=0 it runs off to +∞
