Let `E= x(2^x - 1)(3^x-9)(x-3)`
Here `2^x-1 =rArr x=0 ` and when `3^x-9=0 rArr x=2`
Now mark x=0 , 2 and 3 on real, number line
The sing of E starts with a positive sign from right hand side.
Also at x=0 two factors vanish x and `2^x-1` hence the sing of E dose not change while crossing x=0
The sign scheme of E is as follows.
From the figure we have `E lt 0 " for" x in (2,3)`
