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Example 7

If f(x) = x2- 2x + 3, x Î R. Find the values of x for which f(x) = f (3x - 1)

Solution :

f(x)    = x2 - 2x + 3 Þ f (3x - 1) = (3x - 1)2 - 2 (3x - 1) + 3

= 9x2 - 6x + 1 - 6x + 2 + 3

= 9x2 - 12 x + 6

\ f(x) = f(3x - 1) gives us

x2 - 2x + 3 = 9x2 - 12x + 6

8x2 - 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

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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 cm3

\ V = (4x3 - 48x2 + 144x) cm3

(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

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