Sol. Let f(x) = (4 –3x 2 )

⇒ f ′ (x) = 9x 2 – 6 x – 4

For maximum or minimum, we must have f ′ (x) = 0

⇒ 9x 2 – 6 x – 4 = 0

⇒ x =

⇒ x = ,

Let x 1 = and x 2 = –

Thus, f ′ (x) = 9

If α > 0, then > – and the changes of signs of f ′ (x) are as shown in fig.

Clearly, x 2 = – is a point of local maximum and x 1 = is a point of local minimum.

If α < 0, then > and the changes in the signs of f ′ (x) are as shown in fig.

In this case x 1 = is a point of local maximum and x 2 = – is a point of local minimum.

The difference between the local maximum and local minimum values.

= |f(x 1 ) – f(x 2 )|

= –

= –

= ( 3 – α 2 ) –

= –

=

=

=

Clearly, it will be least when α =1 and the least value is 32/9.