What are the extrema of #h(x) = 7x^5 - 12x^3 + x#?
Extrema are at x=
Factorising h'(x) and equating it to zero, it would be
The critical points are therefore
For x=-1, h''(x)= -68, hence there would be a maxima at x=-1
for x=1, h''(x)= 68, hence there would be a minima at x=1
for x= #-sqrt (1/35), h''(x) =-0.6761+12.1702=11.4941, hence there would be a minima at this point.