F is increasing if f '(x) = 4x^3<span> - 64x >0, so it is the same of x(x^2 -16)>0, implies x>0 or (x-4)(x+4)>0, so x> + or -4 the answer is </span><span>(-4 , 0) ∪ (4 , infinity)
</span>f is decreasing if f '(x) = 4x^3 - 64x <0, so it is the same of x(x^2 -16)<0, implies x<0 or (x-4)(x+4)<0, so x< + or -4 the answer is (- infinity, -4) ∪ (0 , 4)
<span>the local minimum and maximum </span>f '(x) =0, impolies x=+ or -4, or x=0 or f'(o)=0, and f'(- 4)=f'(4)= 0, M(-4, 0) or M(4, 0) or M(0,0)
<span>inflection points can be found by solving f '' (x)=12x^2 - 64 =0 </span>x=+ or - 4sqrt(3) / 3 so the inflection point is and M(- 4sqrt(3) / 3, f'' ( -4sqrt(3)) (smaller x value), and M(4sqrt(3) / 3, f'' (4sqrt(3)) (larger x value)
f is concave up if f ''>0 it means 12x^2 - 64>0, so the interval is (- infinity, -4sqrt(3) U (4sqrt(3), infinity)
f is concave down if f ''<0 it means 12x^2 - 64<0 so the interval is (-4sqrt(3)) U (4sqrt(3))
