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Sagot :
To graph the function [tex]\( f(x) = \sqrt{x - 2} + 1 \)[/tex] and to identify its domain, let's break down the problem step by step:
### Step 1: Understand the Function
The function [tex]\( f(x) = \sqrt{x - 2} + 1 \)[/tex] is a transformation of the basic square root function [tex]\( \sqrt{x} \)[/tex].
1. Inside the square root: The expression [tex]\( x - 2 \)[/tex] indicates a horizontal shift to the right by 2 units.
2. Outside the square root: The constant [tex]\( +1 \)[/tex] indicates a vertical shift upwards by 1 unit.
### Step 2: Determine the Domain
The domain of a function is the set of all possible x-values for which the function is defined.
For the square root function [tex]\( \sqrt{x - 2} \)[/tex], the quantity under the square root [tex]\( x - 2 \)[/tex] must be non-negative:
[tex]\[ x - 2 \geq 0 \][/tex]
Solving this inequality:
[tex]\[ x \geq 2 \][/tex]
Therefore, the domain of [tex]\( f(x) \)[/tex] in interval notation is:
[tex]\[ [2, \infty) \][/tex]
### Step 3: Create a Table of Values
To make the graph easier, we create a table of values for [tex]\( f(x) = \sqrt{x - 2} + 1 \)[/tex]:
[tex]\[ \begin{array}{c|c} x & f(x) \\ \hline 2 & \sqrt{2 - 2} + 1 = 1 \\ 3 & \sqrt{3 - 2} + 1 = \sqrt{1} + 1 = 2 \\ 4 & \sqrt{4 - 2} + 1 = \sqrt{2} + 1 \approx 2.414 \\ 5 & \sqrt{5 - 2} + 1 = \sqrt{3} + 1 \approx 2.732 \\ 6 & \sqrt{6 - 2} + 1 = \sqrt{4} + 1 = 3 \\ 7 & \sqrt{7 - 2} + 1 = \sqrt{5} + 1 \approx 3.236 \\ \end{array} \][/tex]
### Step 4: Plot the Points and Draw the Graph
1. Plot the points [tex]\((2, 1)\)[/tex], [tex]\((3, 2)\)[/tex], [tex]\((4, 2.414)\)[/tex], [tex]\((5, 2.732)\)[/tex], [tex]\((6, 3)\)[/tex], [tex]\((7, 3.236)\)[/tex], etc., on the coordinate plane.
2. Connect these points with a smooth curve which starts at [tex]\((2, 1)\)[/tex] and continues upwards and to the right.
### Step 5: Sketch the Graph
The graph will start at the point [tex]\((2, 1)\)[/tex] and will rise gradually as x increases. The function will increase more slowly as x gets larger because the square root function grows at a decreasing rate.
### Step 6: Identify the Domain in Interval Notation
We previously determined that the domain is:
[tex]\[ [2, \infty) \][/tex]
### Conclusion
- Graph: The graph represents the square root function shifted right by 2 units and up by 1 unit. It starts at the point [tex]\((2, 1)\)[/tex] and moves upwards to the right.
- Domain: The domain in interval notation is [tex]\( [2, \infty) \)[/tex].
This completes the detailed solution and interpretation of the function [tex]\( f(x) = \sqrt{x - 2} + 1 \)[/tex].
### Step 1: Understand the Function
The function [tex]\( f(x) = \sqrt{x - 2} + 1 \)[/tex] is a transformation of the basic square root function [tex]\( \sqrt{x} \)[/tex].
1. Inside the square root: The expression [tex]\( x - 2 \)[/tex] indicates a horizontal shift to the right by 2 units.
2. Outside the square root: The constant [tex]\( +1 \)[/tex] indicates a vertical shift upwards by 1 unit.
### Step 2: Determine the Domain
The domain of a function is the set of all possible x-values for which the function is defined.
For the square root function [tex]\( \sqrt{x - 2} \)[/tex], the quantity under the square root [tex]\( x - 2 \)[/tex] must be non-negative:
[tex]\[ x - 2 \geq 0 \][/tex]
Solving this inequality:
[tex]\[ x \geq 2 \][/tex]
Therefore, the domain of [tex]\( f(x) \)[/tex] in interval notation is:
[tex]\[ [2, \infty) \][/tex]
### Step 3: Create a Table of Values
To make the graph easier, we create a table of values for [tex]\( f(x) = \sqrt{x - 2} + 1 \)[/tex]:
[tex]\[ \begin{array}{c|c} x & f(x) \\ \hline 2 & \sqrt{2 - 2} + 1 = 1 \\ 3 & \sqrt{3 - 2} + 1 = \sqrt{1} + 1 = 2 \\ 4 & \sqrt{4 - 2} + 1 = \sqrt{2} + 1 \approx 2.414 \\ 5 & \sqrt{5 - 2} + 1 = \sqrt{3} + 1 \approx 2.732 \\ 6 & \sqrt{6 - 2} + 1 = \sqrt{4} + 1 = 3 \\ 7 & \sqrt{7 - 2} + 1 = \sqrt{5} + 1 \approx 3.236 \\ \end{array} \][/tex]
### Step 4: Plot the Points and Draw the Graph
1. Plot the points [tex]\((2, 1)\)[/tex], [tex]\((3, 2)\)[/tex], [tex]\((4, 2.414)\)[/tex], [tex]\((5, 2.732)\)[/tex], [tex]\((6, 3)\)[/tex], [tex]\((7, 3.236)\)[/tex], etc., on the coordinate plane.
2. Connect these points with a smooth curve which starts at [tex]\((2, 1)\)[/tex] and continues upwards and to the right.
### Step 5: Sketch the Graph
The graph will start at the point [tex]\((2, 1)\)[/tex] and will rise gradually as x increases. The function will increase more slowly as x gets larger because the square root function grows at a decreasing rate.
### Step 6: Identify the Domain in Interval Notation
We previously determined that the domain is:
[tex]\[ [2, \infty) \][/tex]
### Conclusion
- Graph: The graph represents the square root function shifted right by 2 units and up by 1 unit. It starts at the point [tex]\((2, 1)\)[/tex] and moves upwards to the right.
- Domain: The domain in interval notation is [tex]\( [2, \infty) \)[/tex].
This completes the detailed solution and interpretation of the function [tex]\( f(x) = \sqrt{x - 2} + 1 \)[/tex].
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