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 Now
f ″(x) = 36x2 + 24x – 24 = 12 (3x2 + 2x – 2)
or
′′
= −
<
′′
=
>
′′ −
=
>



f
f
f
( )
( )
(
)
0
24
0
1
36
0
2
72
0
Therefore, by second derivative test, x  = 0 is a point of local maxima and local
maximum value of f at x = 0 is f (0) = 12 while x = 1 and x = – 2 are the points of local
minima and local minimum values of f at x = – 1 and – 2 are f (1) = 7 and f (–2) = –20,
respectively