What is the minima of f(x) = 9x^3 + 6x^2 + 4

To find the the minimum f(x) =
9x^3+6x^2+4

We kow by calculus that f(x) is minimum if f(x)
= 0 for some x= x' and f''(x') < 0.

Therefore
(9x^3+6x^2+4)' = 0 gives:

9*3x^2+6*2x+0 =
0

27x^2+12x = 0

3x(9x+ 4) =
0

x = 0 or x = -4/9

Now  we
find the 2nd differential coefficient:

f"(x) = (27x^2+12x)
= 27*2x+12 = 54x+12

f"(0) = 12 which is > 0. and
f"(-4/9) =  54(-4/9)+12 =  --12

So, f(0) = 9x^3+6x^2+4 = 
9*0^3+6*0^2+4 = 4 is the minimum locally.

But the global
minimum  does not exist as f(x) becomes -infinity as x-->
-infinity.