# Understanding Domain and Range Part 3

In the previous post, we have learned how to analyze equations of functions and determine their domain and range. We have observed that the range of the functions $y = x^2$ and $y = |x|$ are the set of real numbers greater than or equal to $0$ since squaring a number or getting its absolute value results to $0$ or a positive real number. We also learned the value under the square root sign must be greater than or equal to 0. Lastly, we have learned that we cannot divide by zero because it will make the function undefined.

In this post, we are going to continue our discussion by examining functions with equations more complicated than those in the second part of this series.

Squares and Absolute Values

1. $f(x) = x^2 - 3$

Domain: The function is defined for any real number $x$, so the domain of $f$ is the set of real numbers.

Range: The minimum value of $x^2$ is $0$ for any real number $x$ and $f(0) - 3 = 0^2 - 3 = -3$. So, the minimum value of the function is $-3$. We can make the value of the function as large as possible by increasing the absolute value of $x$. So, the range of the function is the set of real numbers greater than or equal to $-3$ or $[-3, \infty)$ in interval notation.

2. $g(x) = |x - 3| + 4$

Domain: The function is defined for any real number $x$, so the domain of $f$ is the set of real numbers.

Range: The minimum value of any number inside the absolute value sign is $0$ (here, that happens at $g(3)$). So adding $4$ to $g(3)$ will make the minimum value of the range $g(3) + 0 = 0 + 4 = 4$. So the range is the set of real numbers greater than or equal to $4$ or $[4, \infty)$ in interval notation.

Division by 0

3. $h(x) = \frac{1}{x-3}$

Domain: The denominator of the equation of the function $h$ must not equal to $0$, so we find the value that will make it $0$. That is, we equate $x - 3$ to $0$. Now, $x - 3 = 0$ gives us $x = 3$. Therefore, $x$ must not equal to $3$. So, the domain of the function $h$ is the set of real numbers except $x \neq 3$ or $(- \infty, 3) \cup (3, \infty)$.  Notice, that to get the restrictions in the domain we equate the denominator to $0$ and solve for $x$. The value of $x$ is the restriction in the domain.

Range: Just like the function $y = 1/x$ in Part 2, we can have any value for $h(x)$, but it’s impossible for it to be $0$ no matter how small $x-3$ is. So, the range is the set of real numbers except 0 or $(- \infty, 0) \cup (0, \infty)$

4. $p(x) = \frac{5}{2x + 5}$

Domain: The denominator of the equation of the function $p$ must not equal to $0$, so we find the value that will make it $0$. That is, we equate $2x + 5$ to $0$. Now, $2x + 5 = 0$ means $x = -5/2$. Therefore, $x$ must not equal $-5/2$. So, the domain of the function is the set of real numbers except $x = -5/2$ or $(- \infty, -5/2) \cup (-5/2, \infty)$.

Range: Left as an exercise

5. $q(x) = \sqrt {x - 5}$

Domain: We have learned that the value inside the radical must be positive or 0, so,
$\sqrt {x-5} \geq 0$

Squaring both sides,
$x - 5 \geq 0$
$x \geq 5$.

So, the domain of the function $q$ is the set of real numbers greater than or equal to 5. That is $[5, \infty)$  in interval notation.

Range: The value inside the radical sign must be greater than or equal to 0, so the range is the set of real numbers greater than or equal to 0.

6. $r(x)= \sqrt {2x + 7}$

Domain:
$\sqrt{ 2x + 7} \geq 0$

Squaring both sides,

$2x + 7 \geq 0$
$2x \geq -7$
$x \geq -7/2$.

So, the domain is the set of real numbers greater than or equal to -7/2 or $[-7/2, \infty)$ in interval notation.

Range: We know that $r(x) = \sqrt {2x - 3} \geq 0$, so the range is the set of real numbers greater than or equal to 0 or $[0, \infty)$ in interval notation.

In the next post, we are going to summarize what we have learned in the three posts.