Now to
the left of 0, f (x) = 3 – x and so f ′(x) = – 1 < 0 Also to
Rationalised 2023-24
APPLICATION OF DERIVATIVES
167
the right of 0, f (x) = 3 + x and so  f ′(x) = 1 > 0 Therefore, by first derivative test, x =
0 is a point of  local minima of  f  and local minimum value of  f  is f (0) = 3 Example 20 Find local maximum and local minimum values of the function  f  given by
f (x) = 3x4 + 4x3 – 12x2 + 12
Solution We have
f (x) = 3x4 + 4x3 – 12x2 + 12
or
f ′(x) = 12x3 + 12x2 – 24x = 12x (x – 1) (x + 2)
or
f ′(x) = 0 at x = 0, x = 1 and x = – 2