Questions 1
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
Questions 2
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) 
Solutions to the Above Questions
Solutions to Questions 1
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 , ∞)
Solutions to Questions 2
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) ∪ (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 , ∞) .
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