You need to solve the equation `f'(x) = 0` to find the critical values of the function such that:

f'(x) = 4x^3 + 12x^2 + 8x

You need to solve f'(x) = 0 such that:

`4x^3 + 12x^2 + 8x = 0`

You need to divide by 4 such that:

`x^3 + 3x^2 + 2x = 0`

You need to factor out x such that:

`x(x^2 + 3x + 2) = 0`

`x_1 = 0`

`x^2 + 3x + 2 = 0`

You should use quadratic formula such that:

`x_(2,3) = (-3+-sqrt(9-8))/2`

`x_(2,3) = (-3+-1)/2`

`x_2 = -1 ; x_3 = -2`

You need to select a value, smaller than -2, x=-3, to evaluate `f'(-3)` such that:

`f'(-3) = -3(9-9+2) = -6lt0`

You need to select a value, between -2 and -1, `x = -1.5` , to evaluate `f'(-1.5) ` such that:

`f'(-1.5) = -1.5(2.25- 4.5 + 2) = -1.5*(-0.25) gt 0`

You need to select a value, between -1 and 0, `x = -0.5` , to evaluate `f'(-0.5)` such that:

`f'(-0.5) = -0.5(0.25 - 1.5 + 2) = -0.5*0.75lt0`

You need to select a value, larger than 0, `x=1` , to evaluate `f'(1) ` such that:

`f'(1) = 1(1+3+2) = 6gt0`

**Hence, the function has local maximum at x=-1 and two local minima
at x=-2 and x=0 and the function decreases over intervals
`(-oo,-2)U(-1,0)` and it increases over intervals
`(-2,-1)U(0,oo).`**

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