Example 7
If f(x) = x^{2} 2x + 3, x Î
R. Find the values of x for which f(x) = f (3x  1)
Solution :
f(x) = x^{2}  2x + 3 Þ f (3x  1) = (3x  1)^{2}  2 (3x  1) + 3
= 9x^{2 } 6x + 1  6x + 2 + 3
= 9x^{2 } 12 x + 6
\ f(x) = f(3x  1) gives us
x^{2}  2x + 3 = 9x^{2}  12x + 6
8x^{2}  10x + 3 = 0
(4x  3) (2x  1) = 0 ...(factorizing)
4x  3 = 0 or 2x  1 = 0
\ x = 3/4 or x = 1/2
\ The required values of x are 3/4 and 1/2
Example 8 An open box is to be made from a rectangular piece of tin 12 cm ´ 12 cm, by cutting four squares from each corner and turning up the sides
(1) Find the formula that expresses its volume
(2) Find the domain of the function expressing its volume
Solution :
(1) Let x cm be the length of the each side of this square to be cut off and V be the volume of the box. Then the dimensions of the open box are as  Length = (12  2x), Breadth = (12  2x) and height = x in cm.
\ volume of the box (V) = Length ´ Breadth ´ Height
\ V = (12  2x) . (12  2x) . x cm^{3}
\ V = (4x^{3}  48x^{2} + 144x) cm^{3}
(2) put v = 0 Þ (12  2x) (12  2x) = 0 Þ x ,= 0 or x = 6
(cut points)
Interval 
Test values 
Values of f(x) i.e. V 
Sign of f(x) i.e. V 
( ¥ , 0) 
1 
(10) (10) (1) =  100 
 
(0, 6) 
5 
(2) (2) (5) = 20 
+ 
(6, ¥) 
7 
( 2) (2) (7) = 28 
+ 
Since the interval (6, 8) is practically impossible (say absurd), the only acceptable interval is (0, 6) i.e. the domain of V is 0 < x < 6
