2011-03-23, 10:38 PM
f(x) = x^3 * x^-1 (x != 0)
f'(x) = (3x^2 * x^-1) + (x^3 * -1x^-2) = (3x^2)/(x) - (x^3/x^2) <-- product rule
f'(x) = 0 <=> (3x^2)/(x) - (x^3/x^2) = 0 <--- bullshit distribution
<=> x^2/x * (3 - x/x) = 0
<=> x^2 = 0 or 3 - x/x = 0
<=> x = 0, or x(3-1)/x = 0
<=> x = 0, or x = 0, or 3-1 = 0
So no, it doesn't have a minimum.
f'(x) = (3x^2 * x^-1) + (x^3 * -1x^-2) = (3x^2)/(x) - (x^3/x^2) <-- product rule
f'(x) = 0 <=> (3x^2)/(x) - (x^3/x^2) = 0 <--- bullshit distribution
<=> x^2/x * (3 - x/x) = 0
<=> x^2 = 0 or 3 - x/x = 0
<=> x = 0, or x(3-1)/x = 0
<=> x = 0, or x = 0, or 3-1 = 0
So no, it doesn't have a minimum.