If f x = x - 1 x , find the value of f f 1 x .

Given, f x = x - 1 x ⇒ f x = 1 - 1 x To find f 1 x we will replace x by 1 x f 1 x = 1 - 1 1 x ' ⇒ f 1 x = 1 - x Now, f f 1 x = f 1 - x = 1 - x - 1 1 - x ⇒ f f 1 x = - x 1 - x

If f x = x - 1 x , find the value of f f 1 x .

Given, f x = x - 1 x ⇒ f x = 1 - 1 x To find f 1 x we will replace x by 1 x f 1 x = 1 - 1 1 x ' ⇒ f 1 x = 1 - x Now, f f 1 x = f 1 - x = 1 - x - 1 1 - x ⇒ f f 1 x = - x 1 - x

Find the range of the function, f x = x x

Given f ( x ) = x | x | Let y = x | x | For x > 0 , | x | = x ⇒ y = x x = 1 For x < 0 , x = - x ⇒ y = x - x = - 1 Thus, the range of f ( x ) is { - 1 , 1 }

If f x + 1 x = x 2 + 1 x 2 for all x ∈ R - 0 then write an expression for f x .

Given f x + 1 x = x 2 + 1 x 2 f x + 1 x = x 2 + 1 x 2 + 2 - 2 [adding and subtracting 2 ] f x + 1 x = x 2 + 1 x 2 + 2 · x · 1 x - 2 f x + 1 x = x + 1 x 2 - 2 [ ∵ ( a + b ) 2 = a 2 + b 2 + 2 a b ] Thus, f x + 1 x = x + 1 x 2 - 2 Hence, Replace x + 1 x with x , f ( x ) = x 2 - 2 Where | x | ≥ 2 .

Write the domain and the range of the function, f x = a x + b b x - a .

Given, f x = a x + b b x - a we know that function is not defined if denominator is zero. ∴ f ( x ) will be defined if b x - a ≠ 0 ⇒ x ≠ a b Therefore, the domain of given function is D f = R - a b . Let f x = y = a x + b b x - a ⇒ b x y - a y = a x + b ⇒ b x y - a x = a y + b ⇒ x = a y + b b y - a The above function is not defined when, b y - a = 0 ⇒ y = a b ∴ y ∈ R - a b Therefore, the range of function is R f = R - a b .