To determine the local extrema of a function, we'll have to find out first if the function does have any critical values.

The critical values are the roots of the first derivative. We'll differentiate the given function with respect to x, to determine the 1st derivative of the function.

f'(x) = 4x^3 - 18x

Now, we'll cancel the 1st derivative to determine it's zeroes:

4x^3 - 18x = 0

2x(2x^2 - 9) = 0

We'll cancel each factor:

2x = 0 => x1 = 0

2x^2 - 9 = 0 => 2x^2 = 9

x^2 = 9/2

x2 = 3/sqrt2

x2 = 3sqrt2/2

x3 = -3sqrt2/2

We'll determine the monotony of the function over the intervals (-`oo`; -3sqrt2/2 ) ; (-3sqrt2/2 ; 0) ; (0;3sqrt2/2) ; (3sqrt2/2 ; `oo` ).

The function is decreasing over the interval (-`oo` ; -3sqrt2/2) and then it increases over (-3sqrt2/2 ; 0), therefore the point f(-3sqrt2/2) represents a minimum local point.

The function is increasing over the interval ( -3sqrt2/2 ; 0) and then it decreases over (0 ; 3sqrt2/2 ), therefore the point f(0) represents a maximum local point.

The function is decreasing over the interval (0 ; 3sqrt2/2 ) and then it increases over (3sqrt2/2 ; `oo` ), therefore the point f(3sqrt2/2) represents a minimum local point.

**Therefore, the local minima of the function are represented by the values f(-3sqrt2/2) and f(3sqrt2/2) and the local maxima is represented by the value f(0).**