Question

7·1 domain and range review worksheet
state the domain and range of each relation. state if the relation is a function.
a) (-5,3),(-3,0),(0,1),(2,3),(3,-3)
b) mapping diagram image
c) circle graph image scatter - plot graph image
) 2x + 5
f) $y = 2(x - 3)^2 + 1$
g) $x^2 + y^2 = 16$
h) $y = 2\sqrt{x + 1}$
handwritten notes on domain, range, function status

Explanation:

Let's take part (h): \( y = 2\sqrt{x + 1} \) to find domain and range.

Step 1: Find the Domain

The expression under the square root (radicand) must be non - negative. So we set up the inequality:
\( x+1\geq0 \)
To solve for \( x \), we subtract 1 from both sides of the inequality:
\( x\geq - 1 \)
So the domain is all real numbers \( x \) such that \( x\geq - 1 \), in set notation: \( \{x\in\mathbb{R}|x\geq - 1\} \)

Step 2: Find the Range

We know that the square root function \( \sqrt{u} \) (where \( u=x + 1\)) has a range of \( \sqrt{u}\geq0 \) for \( u\geq0 \).
We have \( y = 2\sqrt{x + 1} \). Since \( \sqrt{x + 1}\geq0 \), when we multiply both sides of the inequality by 2 (and since 2 is a positive number, the direction of the inequality remains the same), we get:
\( 2\sqrt{x + 1}\geq0 \)
So \( y\geq0 \). The range is all real numbers \( y \) such that \( y\geq0 \), in set notation: \( \{y\in\mathbb{R}|y\geq0\} \)

Answer:

• Domain of \( y = 2\sqrt{x + 1} \): \( \{x\in\mathbb{R}|x\geq - 1\} \)
• Range of \( y = 2\sqrt{x + 1} \): \( \{y\in\mathbb{R}|y\geq0\} \)