College Algebra Questions With Answers
sample 5 : Domain and Range of Functions

College algebra questions on finding the domain and range of functions with answers, are presented. The solutions are at the bottom of the page.

Questions 1

Find the domain.

Domain is found by setting x² - 4 ≠ 0 because division by 0 is not allowed.
Domain: (-∞ , -2) U (-2 , 2) U (2 , ∞)
x² + 4x + 3 ≠ 0 , division by 0 is not allowed.
Solve: x² + 4x + 3 = 0 , solutions: x = -3 and x = -1
Domain: (-∞ , -3) U (-3 , -1) U (-1 , ∞)
x² + 5x - 6 ≥ 0 , quantity under square root has to be positive or equal to zero for function to be real.
Solve: x² + 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) = -x² + 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) = |x³ + 4|
x³ + 4 has a range given by the interval (-∞ , ∞)
Because of the absolute value function k(x) = |x³ + 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) ∪ (0 , ∞). 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 l(x) = 1 / |(x - 3)| has the same range as 1 / |x| which is (0 , ∞) .