f(x) = 2x^4 - x^3
f'(x) = 8x^3 - 3x^2
Setting the derivative to zero, we have:
f'(x) = 0
8x^3 - 3x^2 = 0
x^2(8x - 3) = 0
x^2 = 0 or 8x - 3 = 0
x = 0 or x = 3 / 8
Thus, the critical values we have are x = 0 and x = 3 / 8.
Now, we will draw a number line with 0 and 3/8 on it.
By random substitution, we will realise fo any number less than 0, f'(x) will be more than zero. Similarly, for any number more than 3/8, f'(x) will also be more than zero.
This implies that the function is increasing on (-infinity, 0) and (3/8, infinity).
However, by using the similar random substitution, we will realise that for any value between 0 and 3/8, f'(x) will be negative. Thus, the function is decreasing on the interval (0, 3/8).