Find the domain and the range of the function f x = 4 + x 4 - x .

Domain: Given, f x = 4 + x 4 - x For the domain of a function D f , f x must be a real number. ⇒ 4 + x 4 - x must be a real number ⇒ 4 - x ≠ 0 ⇒ x ≠ 4 ⇒ D f = set of all real numbers except at x = 4 Hence, D f = R - 4 . Range: For R f , let y = 4 + x 4 - x ⇒ - x y + 4 y = 4 + x ⇒ x y + x = 4 y - 4 ⇒ x ( y + 1 ) = 4 y - 4 ⇒ x = 4 y - 4 y + 1 but x ∈ R ⇒ 4 y - 4 y + 1 must be a real number ⇒ y + 1 ≠ 0 ⇒ y ≠ - 1 ⇒ R f = set of all real numbers except at y = - 1 Hence, the range of the function R f = R - - 1 .

Find the domain and the range of the function f x = 4 + x 4 - x .

Domain: Given, f x = 4 + x 4 - x For the domain of a function D f , f x must be a real number. ⇒ 4 + x 4 - x must be a real number ⇒ 4 - x ≠ 0 ⇒ x ≠ 4 ⇒ D f = set of all real numbers except at x = 4 Hence, D f = R - 4 . Range: For R f , let y = 4 + x 4 - x ⇒ - x y + 4 y = 4 + x ⇒ x y + x = 4 y - 4 ⇒ x ( y + 1 ) = 4 y - 4 ⇒ x = 4 y - 4 y + 1 but x ∈ R ⇒ 4 y - 4 y + 1 must be a real number ⇒ y + 1 ≠ 0 ⇒ y ≠ - 1 ⇒ R f = set of all real numbers except at y = - 1 Hence, the range of the function R f = R - - 1 .

Find the domain and the range of the function f x = 16 - x 2 .

Given, f x = 16 - x 2 Domain : For the domain of a function D f , f x must be a real number. ⇒ 16 - x 2 must be a real number ⇒ 16 - x 2 ≥ 0 (The square root of a number is always non negative) ⇒ x 2 - 16 ≤ 0 ⇒ x - 4 x + 4 ≤ 0 We know that, if x - a x + a ≤ 0 then x ∈ - a , a . Hence, D f = [ - 4 , 4 ] . Range : Since, - 4 ≤ x ≤ 4 ⇒ 0 ≤ x 2 ≤ 16 (Square of any number is always non-negative) ⇒ 0 ≥ - x 2 ≥ - 16 (Multiplied by - 1 on each side of inequality so that, the inequality gets reversed) ⇒ 16 + 0 ≥ 16 - x 2 ≥ 16 - 16 (Added 16 on each side of inequality) ⇒ 16 ≥ 16 - x 2 ≥ 0 ⇒ 16 ≥ 16 - x 2 ≥ 0 ⇒ 4 ≥ f x ≥ 0 Hence, the range of the function is R f = 0 , 4 .

Find the domain and the range of the function f x = x 2 - 9 .

Given, f x = x 2 - 9 Domain : For the domain of a function D f , f x must be a real number. ⇒ x 2 - 9 must be a real number ⇒ x 2 - 9 ≥ 0 (The square root of a number is always non negative) ⇒ x - 3 x + 3 ≥ 0 We know that, if x - a x + a ≥ 0 then x ∈ ( - ∞ - a ] ∪ [ a , ∞ ) . Hence, D f = ( - ∞ , - 3 ] ∪ [ 3 , ∞ ) . Range : Let y = x 2 - 9 → 1 As square root of a real number is always non-negative, y ≥ 0 On squaring 1 , we get ⇒ y 2 = x 2 - 9 ⇒ x 2 = y 2 + 9 but x 2 ≥ 0 , for all x ∈ R ⇒ y 2 + 9 ≥ 0 ⇒ y 2 ≥ - 9 , which is true for all y ∈ R also , y ≥ 0 Hence, the range of the function is R f = [ 0 , ∞ ) .

Find the domain and the range of the function f x = 1 4 - x 2 .

Given, f x = 1 4 - x 2 Domain : For the domain of a function D f , f x must be a real number. ⇒ 1 4 - x 2 must be a real number ⇒ 4 - x 2 > 0 (The square root of a number is always non negative) ⇒ x 2 - 4 < 0 (Multiplied by - 1 on both sides) ⇒ x - 2 x + 2 < 0 We know that, if x - a x + a < 0 then x ∈ - a , a . Hence, D f = ( - 2 , 2 ) . Range : lim x → 2 - 1 4 - x 2 = 1 0 + = ∞ and lim x → - 2 + 1 4 - x 2 = 1 0 + = ∞ When x = 0 f ( 0 ) = 1 4 = 1 2 The range is [ 1 2 , ∞ ]