Find the domain and range of the real function f ( x ) = x - 3 .

Let y = f ( x ) = x - 3 Domain : For the domain of a function D f , f x must be a real number. ⇒ x - 3 must be a real number ⇒ x - 3 ≥ 0 (Square root never be negative) ⇒ x ≥ 3 Since we know If x ≥ a then x ∈ [ a , ∞ ) Hence, D f = [ 3 , ∞ ) . Range : Since, x ≥ 3 ⇒ x - 3 ≥ 3 - 3 (Added - 3 on both side of inequality) ⇒ x - 3 ≥ 0 ⇒ x - 3 ≥ 0 (Since square root is always positive) ⇒ y ≥ 0 Given Hence, the range of the function is [ 0 , ∞ ) .

Find the domain and the range of the function f x = 25 - x 2 .

Given, f x = 25 - x 2 Domain : For the domain of a function D f , f x must be a real number. ⇒ 25 - x 2 must be a real number ⇒ 25 - x 2 ≥ 0 (The square root of a number is always non -negative) ⇒ x 2 - 25 ≤ 0 ⇒ x - 5 x + 5 ≤ 0 We know that, if x - a x + a ≤ 0 then x ∈ - a , a . Hence, D f = [ - 5 , 5 ] . Range : Since, - 5 ≤ x ≤ 5 ⇒ 0 ≤ x 2 ≤ 25 ⇒ 0 ≥ - x 2 ≥ - 25 ⇒ 25 + 0 ≥ 25 - x 2 ≥ 25 - 25 ⇒ 25 ≥ 25 - x 2 ≥ 0 ⇒ 25 ≥ 25 - x 2 ≥ 0 ⇒ 5 ≥ f x ≥ 0 Hence, the range of the function is R f = 0 , 5 . (Minimum to maximum value).

Draw the graph of the real function f x = 1 + 2 x , x < 0 3 + 5 x , x ≥ 0 . Hence find their range.

Given f x = 1 + 2 x , x < 0 3 + 5 x , x ≥ 0 Since the function is defined for every x ∈ R . Hence, the domain of the function is D f = R Let y = f x = 1 + 2 x , x < 0 3 + 5 x , x ≥ 0 , When x < 0 , y = 1 + 2 x Which is the first degree equation in x , y and hence it represents a straight line. Two points are sufficient to determine a straight line uniquely. The table of values are x - 1 - 2 y - 1 - 3 The graph is a part of a straight line. When x ≥ 0 , y = 3 + 5 x , which is also a first degree equation in x , y The table of values are x 0 1 y 3 8 The graph is part of a straight line. The graph of the function is shown in figure. From the graph it is cleat that, the range of function is ( - ∞ , 1 ) ∪ [ 3 , ∞ ) .