a) You need to perform the first derivative test such that:

`f'(x) = 8x - 4x^3`

You should solve the equation f'(x) = 0 such that:

`8x - 4x^3 = 0 =gt 2x - x^3 = 0`

`x(2 - x^2) = 0`

`x = 0`

`x^2 = 2 =gt x_(1,2) =+-sqrt2`

You need to select a value between `-sqrt2 ` and 0 such that:

`x = -1 =gt f'(-1) = - 8 + 4 = -4 lt 0`

You need to select a value between 0 and `sqrt 2` such that:

`x = 1 =gt f'(1) = 8 - 4 = 4 gt 0`

You need to select a value larger than `sqrt 2` such that:

`f(2) = 16 - 32 = -16 lt 0`

You need to select a value smaller than `-sqrt 2` such that:

`f(-2) = -16 + 32 = 16 gt 0`

Hence, the function increases where derivative is positive and it decreases where derivative is negative.

**Hence, the function increases over `(-oo,-sqrt2)U(0,sqrt2)` and it decreases over `(-sqrt2,0)U(sqrt2,oo).` **

b)** Hence, the function reaches its maxima at x = `-sqrt2 ` and x=`sqrt2` and it reaches its minimum at x=0.**

c)You need to solve the equation f''(x)=0 to find inflection points such that:

`f''(x) = 8 -12x^2`

`8 -12x^2 = 0 =gt 2 - 3x^2 = 0 =gt x^2 = 2/3 =gt x_(1,2) = +-sqrt6/3`

**Hence, the inflection points of the function have x coordinate at `x=-sqrt6/3` and `x=sqrt6/3` .**

d) The function is concave up if f"(x)>0 and it is concave down if f''(x) < 0.

Using the signs law of quadratics yields that f''(x)>0 over interval `(-sqrt6/3 ; sqrt6/3)` and f''(x)<0 over `(-oo;-sqrt6/3)U(sqrt6/3;oo).`

**Hence, the function is concave up if `x in (-sqrt6/3 ; sqrt6/3)` and it is concave down if `x in (-oo;-sqrt6/3)U(sqrt6/3;oo).` **

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