f(x)=-2x^3+30x^2-144x+6. What are the 2 critcal numbers A & B, then find f"(A)= and f"(B)=. where are the local max and min? 

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You need to determine the critical values A and B, hence, you need to solve for x the equation f'(x) = 0, such that:

`f'(x) = (-2x^3 + 30x^2 - 144x + 6)'`

`f'(x) = -6x^2 + 60x - 144`

You need to solve the equation f'(x) = 0, hence, you need to substitute -`6x^2 + 60x - 144` for `f'(x)` , such that:

`-6x^2 + 60x - 144 = 0`

Factoring out by -6 yields:

`-6(x^2 - 10x + 24) = 0`

Dividing by -6 yields:

`x^2 - 10x + 24 = 0`

You may substitute -`10 x` with `-4x - 6x` , such that:

`x^2 -4x - 6x + 24 = 0`

Grouping the terms, yields:

`(x^2 -4x) - (6x - 24) = 0`

Factoring out x in the first group and 6 in the second group yields:

`x(x - 4) - 6(x - 4) = 0 => (x - 4)(x - 6) = 0`

`x - 4 = 0 => x_1 = 4`

`x - 6= 0 => x_2 = 6`

Hence, evaluating the critical values yields `A = 4` and `B = 6.`

You need to evaluate the second order derivative, such that:

`f''(x) = (f'(x))' => f''(x) = (-6x^2 + 60x - 144)'`

`f''(x) = -12x + 60`

You need to evaluate `f''(A)` and `f''(B)` , such that:

`f''(A) = f''(4) => f''(A) = -12*A + 60 => f''(4) = -12*4 + 60 => f''(4) = 60 - 48 = 12`

`f''(B) = f''(6) => f''(B) = -12*B + 6 => f''(6) = -12*6 + 60 = -12`

Hence, evaluating f''(A) and `f''(B)` , under the given conditions, yields `f''(A) = f''(4) = 12` and` f''(B) = f''(6) = -12.`

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