Solution We have
f (x) = 2x3 – 15x2 + 36x + 1
or
f ′(x) = 6x2 – 30x + 36 = 6 (x – 3) (x – 2)
Note that f ′(x) = 0 gives  x = 2 and x = 3 We shall now evaluate the value of f at these points and at the end points of the
interval [1, 5], i e , at x = 1, x = 2, x = 3 and at x = 5