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))
First find out how much one onion weighs so divide 0.75 by 3. This gets you 0.25. Now all you need to do is multiply the weight of one onion by 10, 0.25 x 10 = 2.5
