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F(x) = 3 x⁴ + 4x³ - 12 x² + 12

find derivative:

f '(x) = 12 x³ + 12 x² - 24 x = 0 =>

x = 0 or x² + x - 2 = 0

=> (x +2) (x -1) = 0

So local maximum of minimum occur at x = -2, 0, 1.

Find the second derivative.

f ' '(x) = 36 x² + 24 x - 24 = 12 (3x² + 2 x - 2)

f ' '(-2) = 72 , f ' '(0) = -24 , f ' '(1) = 36

So there is a local minimum at x = -2, a local maximum at x = 0 and a local minimum at x = 1. Local minimum if second derivative is positive. local maximum when second derivative is negative.

f(x) = 3x⁴ + 4x³ -12 x² + 12

f(-3) = 39 ,

find derivative:

f '(x) = 12 x³ + 12 x² - 24 x = 0 =>

x = 0 or x² + x - 2 = 0

=> (x +2) (x -1) = 0

So local maximum of minimum occur at x = -2, 0, 1.

Find the second derivative.

f ' '(x) = 36 x² + 24 x - 24 = 12 (3x² + 2 x - 2)

f ' '(-2) = 72 , f ' '(0) = -24 , f ' '(1) = 36

So there is a local minimum at x = -2, a local maximum at x = 0 and a local minimum at x = 1. Local minimum if second derivative is positive. local maximum when second derivative is negative.

f(x) = 3x⁴ + 4x³ -12 x² + 12

f(-3) = 39 ,

**f(-1) = -1 ,***f(-2) = -30 ,***,***f(0) = 12***, f(2) = 44***f(1) = 7*