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The problem in the image asks to find the range ("область значень") of the function \( y = x + 3 \), given different domains of \( x \).

Let's go through each part:

### Function:
\[ y = x + 3 \]

### Part (a): Domain \( [-3; 3] \)

If \( x \) is between \(-3\) and \(3\), we substitute these values into the function:

- When \( x = -3 \), \( y = -3 + 3 = 0 \)
- When \( x = 3 \), \( y = 3 + 3 = 6 \)

Thus, the range is \( [0; 6] \).

### Part (b): Domain \( [1; 7] \)

For \( x \) between \(1\) and \(7\), we calculate:

- When \( x = 1 \), \( y = 1 + 3 = 4 \)
- When \( x = 7 \), \( y = 7 + 3 = 10 \)

So, the range is \( [4; 10] \).

### Part (c): Domain \( [0; +\infty) \)

For \( x \geq 0 \), the function becomes:

- When \( x = 0 \), \( y = 0 + 3 = 3 \)
- As \( x \) increases, \( y \) increases without bound.

Thus, the range is \( [3; +\infty) \).

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Would you like more details on any part of the solution?

Here are 5 related questions to consider:
1. What would the range be if the domain was \( (-\infty; 0] \)?
2. How does shifting the function \( y = x + 3 \) affect the range compared to \( y = x \)?
3. What happens to the range if we change the function to \( y = x - 3 \)?
4. Can the range of this linear function ever be restricted to a finite interval?
5. How does the slope of the function \( y = x + 3 \) influence the range?

**Tip:** For linear functions like \( y = mx + b \), the slope \( m \) and the intercept \( b \) define how the function behaves over any domain, making range calculations straightforward.

Знайдіть область значень функції y = x + 3, заданої на проміжку: a) [-3; 3]; b) [1; 7]; в) [0; +∞).

Learn how to find the range of the linear function y = x + 3 over different domains. This problem demonstrates how to calculate the range for linear functions on intervals like [-3; 3], [1; 7], and [0; +∞). Suitable for students in Grades 8-10.

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