# 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

### Questions 1

Find the domain of the functions
1. $$f(x) = \dfrac{1}{|x^2 - 4|}$$
2. $$g(x) = \dfrac{1}{x^2 + 4x + 3}$$
3. $$h(x) = \sqrt{x^2 + 5x - 6}$$
4. $$k(x) = \dfrac{1}{\sqrt{(x - 2)^2}}$$
5. $$j(x) = \dfrac{1}{x - \sqrt{x + 2}}$$
6. $$l(x) = \ln{|x + 3|} - 5$$

### Questions 2

Find the range of the functions
1. $$f(x) = -x^2 + 6x + 5$$
2. $$g(x) = |x + 3| - 2$$
3. $$h(x) = \dfrac{x - 2}{x + 3}$$
4. $$k(x) = |x^3 + 4|$$
5. $$j(x) = |(x + 4)(x - 2)|$$
6. $$l(x) = \left| \dfrac{1}{x - 3} \right|$$

## Solutions to the Above Questions

### Solutions to Questions 1

Find the domain.
1. Domain is found by setting $$x^2 - 4 \neq 0$$ because division by 0 is not allowed.
Domain: $$(- \infty, -2) \cup (-2, 2) \cup (2, +\infty)$$
2. $$x^2 + 4x + 3 \neq 0$$, division by 0 is not allowed.
Solve: $$x^2 + 4x + 3 = 0$$, solutions: $$x = -3$$ and $$x = -1$$
Domain: $$(- \infty, -3) \cup (-3, -1) \cup (-1, +\infty)$$
3. $$x^2 + 5x - 6 \geq 0$$, quantity under square root has to be positive or equal to zero for function to be real.
Solve: $$x^2 + 5x - 6 \geq 0$$, solution set: $$(- \infty, -6) \cup (1, +\infty)$$
Domain: $$(- \infty, -6) \cup (1, +\infty)$$
4. $$(x - 2)^2 > 0$$, 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: $$(- \infty, 2) \cup (2, +\infty)$$
5. $$x - \sqrt{x + 2} \neq 0$$, division by 0 not allowed.
Also: $$x + 2 > 0$$, expression under square root has to be positive.
Solve: $$x - \sqrt{x + 2} \neq 0$$, solution: $$x = 2$$
Solve: $$x + 2 > 0$$, solution set: $$[-2, 2) \cup (2, +\infty)$$
6. $$x + 3 > 0$$, argument of logarithm must be positive for function to take real values.
Domain: $$(-3, +\infty)$$

### Solutions to Questions 2

Find the range.
1. $$f(x) = -x^2 + 6x + 5 = -(x - 3)^2 + 14$$
$$-(x - 3)^2 \leq 0$$, for all $$x$$ real.
Add 14 to both sides of the inequality
$$-(x - 3)^2 + 14 \leq 14$$
The left side is the given function. Hence the range is given by the interval
$$(- \infty, 14]$$
2. $$g(x) = |x + 3| - 2$$
$$| x + 3| \geq 0$$, for all $$x$$ real.
Add -2 to both sides of the inequality
$$|x + 3| - 2 \geq -2$$
The left side is the given function. Hence the range is given by the interval
$$[-2, +\infty)$$
3. $$h(x) = \dfrac{x - 2}{x + 3}$$
The inverse function of the given function $$h$$ is
$$h^{-1}(x) = \dfrac{3x + 2}{1 - x}$$
The range of $$h$$ is the domain of $$h^{-1}$$ and is given by the interval $$(- \infty, 1) \cup (1, +\infty)$$
4. $$k(x) = |x^3 + 4|$$
$$x^3 + 4$$ has a range given by the interval $$(- \infty, +\infty)$$
Because of the absolute value function $$k(x) = |x^3 + 4|$$ has a range given by the interval $$[0, +\infty)$$
5. $$j(x) = |(x + 4)(x - 2)|$$
The graph of $$y = (x + 4)(x - 2)$$ 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 $$j(x) = |(x + 4)(x - 2)|$$ is given by the interval $$[0, +\infty)$$
6. $$l(x) = \left| \dfrac{1}{x - 3} \right|$$
The range of $$\dfrac{1}{x}$$ is given by $$(- \infty, 0) \cup (0, +\infty)$$. The range of $$\dfrac{1}{|x|}$$ is given by $$(0, +\infty)$$. The graph of $$\dfrac{1}{|(x - 3)|}$$ is the graphs of $$\dfrac{1}{x}$$ shifted 3 units to the right and therefore $$l(x) = \dfrac{1}{|(x - 3)|}$$ has the same range as $$\dfrac{1}{|x|}$$ which is $$(0, +\infty)$$.