A set of college algebra problems on finding the domain and range of functions with answers, are presented. The solutions are at the bottom of the page.

Find the domain of the functions

f(x) = 1 / x^{ 2}  4

g(x) = 1 / (x^{ 2} + 4x + 3)

h(x) = √(x^{ 2} + 5x  6)

k(x) = 1 / √(x  2)^{ 2}
 j(x) = 1 / (x  √(x + 2))

l(x) = lnx + 3  5

Find the range of the functions

f(x) = x^{ 2} + 6x + 5
 g(x) =  x + 3   2
 h(x) = (x  2) / (x + 3)

k(x) = x^{ 3} + 4
 j(x) =  (x + 4) (x  2) 

l(x) =  1/(x  3) 
Answers to the Above Questions

Find the domain.

Domain is found by setting x^{ 2}  4 ≠ 0 because division by 0 is not allowed.
Domain: (∞ , 2) U (2 , 2) U (2 , ∞)

x^{ 2} + 4x + 3 ≠ 0 , division by 0 is not allowed.
Solve: x^{ 2} + 4x + 3 = 0 , solutions: x = 3 and x = 1
Domain: (∞ , 3) U (3 , 1) U (1 , ∞)

x^{ 2} + 5x  6 ≥ 0 , quantity under square root has to be positive or equal to zero for function to be real.
Solve: x^{ 2} + 5x  6 ≥ 0 , solution set: (∞ , 6) U (1 , ∞)
Domain: (∞ , 6) U (1 , ∞)

(x  2)^{ 2} > 0 , quantity under square root has to be positive. It is a square. It cannot be zero because division by zero is not allowed. Hence x must be different from 2.
Domain: (∞ , 2) U (2 , ∞)

x  √(x + 2) ≠ 0 , division by 0 not allowed.
Also: x + 2 > 0 , expression under square root has to be positive.
Solve: x  √(x + 2) ≠ 0 , solution: x = 2
Solve: x + 2 > 0 , solution set: [2 , 2) U (2 , ∞)

x + 3 > 0 , argument of logarithm must be positive for function to take real values.
Domain: (3 , ∞)

Find the range.

f(x) = x^{ 2} + 6x + 5 = (x  3)^{ 2} + 14
(x  3)^{ 2} ≤ 0 , for all x real.
Add 14 to both sides of the inequality
(x  3)^{ 2} + 14 ≤ 14
The left side is the given function. Hence the range is given by the interval
(∞ , 14]

g(x) = x + 3  2
x + 3 ≥ 0 , for all x real.
Add 2 to both sides of the inequality
x + 3  2 ≥ 2
The left side is the given function. Hence the range is given by the interval
[2 , ∞)

h(x) = (x  2)/(x + 3)
The inverse function of the given function h is
h^{ 1} (x) = (3x + 2) / (1  x)
The range of h is the domain of h^{ 1} and is given by the interval (∞ , 1) U (1 , ∞)

k(x) = x^{ 3} + 4
x^{ 3} + 4 has a range given by the interval (∞ , ∞)
Because of the absolute value function k(x) = x^{ 3} + 4 has a range given by the interval [0 , ∞)

j(x) =  (x + 3) (x  2) 
The graph of the y = (x + 4) (x  2) is a parabola with x intercepts at x = 4 and x = 2 and a vertex at (1 , 9) opening up upward. Part of the graphs in the interval (4 , 2) between the x intercepts is under the x axis and will therefore be reflected above the x axis when the absolute value is applied. Hence the range of j(x) =  (x + 3) (x  2)  is given by the interval [0 , ∞)

l(x) =  1/(x  3) 
The range of 1/x is given by (0 , ∞). The graph of 1/(x  3) is the graphs of 1/x shifted 3 units to the right and therefore 1/(x  3) has the same range as 1/x. Taking the absolute value of 1/(x  3) to get l(x) =  1/(x  3)  will change the range to (0 , ∞) which is the range of l(x).
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