Draw the graph of the following curve : y = - 1 | 2 x + 1 |

Given, y = f x = - 1 | 2 x + 1 | Domain of f x is R - - 1 2 Case- ( i ) : When x < - 1 2 , ⇒ | 2 x + 1 | = - ( 2 x + 1 ) ⇒ y = - 1 - ( 2 x + 1 ) = 1 2 x + 1 = f x For x → - 1 2 - , lim x → - 1 2 - f x = lim h → 0 f - 1 2 - h = lim h → 0 1 2 - 1 2 - h + 1 = lim h → 0 f 1 - 1 - 2 h + 1 = lim h → 0 - 1 2 h = - ∞ So, x = - 1 2 is the vertical asymptote to the graph of f x For x → - ∞ , lim x → - ∞ f x = lim x → - ∞ 1 2 x + 1 = 0 ⇒ x → - ∞ , f x = 0 So, the graph of f x for x < - 1 2 is Case- ( i i ) : When x > - 1 2 , ⇒ 2 x + 1 = 2 x + 1 ⇒ y = - 1 | 2 x + 1 | = - 1 ( 2 x + 1 ) = f x Put x = 0 ⇒ f 0 = - 1 2 ( 0 ) + 1 = - 1 For x → - 1 2 + , lim x → - 1 2 + f x = lim h → 0 f - 1 2 + h = lim h → 0 - 1 2 - 1 2 + h + 1 = lim h → 0 - 1 2 h → - ∞ ⇒ x = - 1 2 is the vertical asymptote to the graph of f x For x → ∞ , lim x → ∞ f x = lim x → ∞ - 1 2 x + 1 = 0 ⇒ x → ∞ , f x → 0 So, the graph of f x for x > - 1 2 is Hence, the graph of y = - 1 | 2 x + 1 | For x ∈ R - - 1 2 is

Draw the graph of the following curve : y | x | - 1 = - 1

Given, y x - 1 = - 1 The domain is x - 1 ≠ 0 ⇒ x ≠ 1 . ⇒ x ≠ 1 , - 1 So, the domain is x ∈ - ∞ , ∞ - - 1 , 1 . As, y x - 1 = - 1 ⇒ y = - x - 1 , x ≠ - 1 , 1 y = - x + 1 So, y ∈ ( - ∞ , 1 ] - 0 . Now, the graph of y = - x is, So, the graph of y = - x + 1 will be (by graphical transformation) The graph of y = - x will be shifted 1 unit upward. At x = 0 , the value of the function is y = 1 .

Draw the graph of the following curve: y = x 2 - 1 x 2 - 1 ln x

Given, y = f x = | x 2 - 1 | ( x 2 - 1 ) ln x Domain of ln x is x > 0 Therefore, the domain of f x is 0 , ∞ - { 1 } Case- ( i ) : x 2 - 1 > 0 ⇒ | x 2 - 1 | = x 2 - 1 ⇒ ( x - 1 ) ( x + 1 ) > 0 (Number line) ⇒ x ∈ ( - ∞ , - 1 ) ∪ ( 1 , ∞ ) Using the domain of f x , x ∈ 1 , ∞ ⇒ For x ∈ ( 1 , ∞ ) , | x 2 - 1 | = + ( x 2 - 1 ) ⇒ f x = | x 2 - 1 | ( x 2 - 1 ) ln x , x ∈ 1 , ∞ f x = ( x 2 - 1 ) ( x 2 - 1 ) ln x = ln x Graph of f ( x ) = ln x for x ∈ ( 1 , ∞ ) is Case- ( i i ) : x 2 - 1 < 0 ⇒ | x 2 - 1 | = - ( x 2 - 1 ) ⇒ ( x - 1 ) ( x + 1 ) < 0 (Number line) ⇒ x ∈ ( - 1 , 1 ) According to the domain of f ( x ) , x ∈ ( 0 , 1 ) ⇒ For x ∈ ( 0 , 1 ) , | x 2 - 1 | = - ( x 2 - 1 ) ⇒ f x = - ( x 2 - 1 ) ( x 2 - 1 ) ln x , x ∈ 0 , 1 = - ln x We knew the graph of y = ln x for x ∈ ( 0 , 1 ) is By graphical transformation, graph of y = - ln x for x ∈ ( 0 , 1 ) is Hence, the graph of f x = | x 2 - 1 | ln x ( x 2 - 1 ) for x ∈ ( 0 , ∞ ) - { 1 } is