Right choice is (c) Does not possess a maximum or minimum value
Easiest explanation: We have, f(x) = 2/3(x^3) – 6x^2 + 20x – 5 ……….(1)
Differentiating both side of (1) with respect to x, we get,
f’(x) = 2x^2 – 12x + 20
Now, for a maximum and minimum value of f(x) we have,
f’(x) = 0
Or 2x^2 – 12x + 20 = 0
Or x^2 – 6x + 10= 0
So, x = [6 ± √(36 – 4*10)]/2
x = (6 ± √-4)/2, which is imaginary.
Hence, f’(x) does not vanishes at any point of x.
Thus, f(x) does not possess a maximum or minimum value.