h(x)=8-|x+4| has a maximum of 8 when x=-4, that is, (-4,8), because the absolute value is always positive and reduces 8 by a positive quantity unless the positive quantity is zero.
f(x)=x2-6x+14, f'(x)=2x-6=0 at extrema, so x=3; f"(3)=2 so there is a minimum at x=3, and f(3)=9-18+14=5. The minimum is (3,5).
m(x)=14+21x-2x2-3x3; m'(x)=21-4x-9x2=0 doesn't factorise rationally.
9x2+4x-21=0, x=(-4±√(16+756))/18=(-4±√772)/18=1.3214 and -1.7658 approx.
m"(x)=-4-18x, m"(1.3214)<0 (max) and m"(-1.7658)>0 (min).
So there's a maximum at (1.3214,31.3353) and a minimum at (-1.7658,-12.8003).