The slope of the tangent at (1, f( 1)) is f' (1) = 3a+b.

Because the tangent is perpendicular to the known straight line, the product of slopes is-1, that is, 3a+b= 18.

The derivative function f' (x) = 3ax2+B. The minimum value is 12, that is, A >；; 0，b= 12。 So a=2.

So a = 2, b = 12 and c = 0. f (x) = 2x 3+ 12x。

Because g(x)=f(x)? X 2, it must be divided.

If g (x) = f (x)/x 2, then g (x) = 2x+12/x > =2√(24)=4√6. The equal sign holds if and only if 2x= 12/x, that is, x=√6. That is, the minimum value of g(x) is 4√6.

If g (x) = f (x) * x 2, then g (x) = 2x 5+12x 3. G' (x) = 10x 4+36x 2 > 0。 Therefore, g(x) monotonically increases at (0, +∞). So there is no minimum in the open interval (0, +∞).