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

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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.

Given:

f(x) = 9x^3 + 6x^2 + 4

To find minima of this function we must find value of x for which f'(x) is equal to 0, and than find value of f(x) for this value of x. We do this as follows.

f'(x) = 9*3x^2 + 6*2x

Simplifying the above expression for f'(x) and equating it to 0, we get:

27x^2 + 12x = 0

3x(9x + 4) = 0

This means that x has two values:

x = 0 and x = - 4/9

Substituting these two value of x in the function f(x):

f(0) = 9*0^3 + 6*0^2 + 4 = 4

f(0) = -9*(4/9)^3 + 6*(4/9)^2 + 4

= - 64/81 + 24/81 + 4

= 192/81

Thus the minimum value of f(x) is 192/81 which occurs at x=-9/4.

Answer:

Minima of f(x) = 192/81

We have the function f(x) = 9 x^3 + 6 x^2 + 4

Now the derivative of the function f'(x) = 27x^2 + 12 x

To find the extreme values we have to equate f'(x) to 0

=> 27x^2 + 12 x =0

=> 3x ( 9x + 4)=0

So x can be 0 or -4/9

Now we take the second derivative of the function

f''(x) = 27x + 12

For x = 0, f''(0)= 12 which is positive, therefore at x= 0 we have a minima.

**So the minimum value of the function is f(0) = 4**