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Sagot :
Let's sketch the graph of the function [tex]\( f(x) = \sin(x) + 2 \)[/tex] for [tex]\( 0^\circ < x < 360^\circ \)[/tex].
### Step-by-Step Solution:
1. Identify the Function Form:
The given function [tex]\( f(x) = \sin(x) + 2 \)[/tex] represents a sine wave that has been vertically shifted upwards by 2 units.
2. Determine Intercepts and Characteristics:
- Y-intercept:
The y-intercept occurs when [tex]\( x = 0 \)[/tex].
[tex]\[ f(0) = \sin(0) + 2 = 0 + 2 = 2 \][/tex]
So, the y-intercept is at [tex]\((0, 2)\)[/tex].
- X-intercepts:
To find the x-intercepts, set [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ \sin(x) + 2 = 0 \quad \Rightarrow \quad \sin(x) = -2 \][/tex]
Since [tex]\( \sin(x) \)[/tex] ranges from [tex]\(-1\)[/tex] to [tex]\(1\)[/tex], there are no x-intercepts as [tex]\( \sin(x) = -2 \)[/tex] is impossible within the given range.
3. Determine the Maximum and Minimum Points:
- The maximum value of [tex]\( \sin(x) \)[/tex] is [tex]\(1\)[/tex]. Therefore, the maximum value of [tex]\( f(x) \)[/tex] is:
[tex]\[ f_{\text{max}} = 1 + 2 = 3 \][/tex]
This occurs at [tex]\( x = 90^\circ \)[/tex] and [tex]\( x = 270^\circ \)[/tex].
- The minimum value of [tex]\( \sin(x) \)[/tex] is [tex]\(-1\)[/tex]. Therefore, the minimum value of [tex]\( f(x) \)[/tex] is:
[tex]\[ f_{\text{min}} = -1 + 2 = 1 \][/tex]
This occurs at [tex]\( x = 180^\circ \)[/tex] and [tex]\( x = 360^\circ \)[/tex].
4. Plot the Function Values:
- At [tex]\( x = 0^\circ \)[/tex], [tex]\( f(0) = 2 \)[/tex]
- At [tex]\( x = 90^\circ \)[/tex], [tex]\( f(90) = 3 \)[/tex]
- At [tex]\( x = 180^\circ \)[/tex], [tex]\( f(180) = 1 \)[/tex]
- At [tex]\( x = 270^\circ \)[/tex], [tex]\( f(270) = 3 \)[/tex]
- At [tex]\( x = 360^\circ \)[/tex], [tex]\( f(360) = 2 \)[/tex]
5. Sketch the Graph:
- Draw the x-axis ranging from [tex]\( 0^\circ \)[/tex] to [tex]\( 360^\circ \)[/tex].
- Draw the y-axis and mark the intercept at [tex]\((0, 2)\)[/tex].
- Plot points at [tex]\((90^\circ, 3)\)[/tex], [tex]\((180^\circ, 1)\)[/tex], [tex]\((270^\circ, 3)\)[/tex], and [tex]\((360^\circ, 2)\)[/tex].
- Sketch a smooth curve through these points to illustrate the sine wave shifted up by 2 units.
### Sketch of [tex]\( f(x) = \sin(x) + 2 \)[/tex]:
```
y
^
3 |
|
|
2 |------------------------------------------------> x
|
1 | *
|
|
0 +------------------------------------------------------
0° 90° 180° 270° 360°
```
In the graph:
- The y-intercept is clearly marked at [tex]\((0, 2)\)[/tex].
- The curve shows one full period of the sine function shifted up by 2 units, peaking at [tex]\(90^\circ\)[/tex] and [tex]\(270^\circ\)[/tex].
The sketch accurately represents [tex]\( f(x) = \sin(x) + 2 \)[/tex] over the interval [tex]\( 0^\circ \)[/tex] to [tex]\( 360^\circ \)[/tex].
### Step-by-Step Solution:
1. Identify the Function Form:
The given function [tex]\( f(x) = \sin(x) + 2 \)[/tex] represents a sine wave that has been vertically shifted upwards by 2 units.
2. Determine Intercepts and Characteristics:
- Y-intercept:
The y-intercept occurs when [tex]\( x = 0 \)[/tex].
[tex]\[ f(0) = \sin(0) + 2 = 0 + 2 = 2 \][/tex]
So, the y-intercept is at [tex]\((0, 2)\)[/tex].
- X-intercepts:
To find the x-intercepts, set [tex]\( f(x) = 0 \)[/tex]:
[tex]\[ \sin(x) + 2 = 0 \quad \Rightarrow \quad \sin(x) = -2 \][/tex]
Since [tex]\( \sin(x) \)[/tex] ranges from [tex]\(-1\)[/tex] to [tex]\(1\)[/tex], there are no x-intercepts as [tex]\( \sin(x) = -2 \)[/tex] is impossible within the given range.
3. Determine the Maximum and Minimum Points:
- The maximum value of [tex]\( \sin(x) \)[/tex] is [tex]\(1\)[/tex]. Therefore, the maximum value of [tex]\( f(x) \)[/tex] is:
[tex]\[ f_{\text{max}} = 1 + 2 = 3 \][/tex]
This occurs at [tex]\( x = 90^\circ \)[/tex] and [tex]\( x = 270^\circ \)[/tex].
- The minimum value of [tex]\( \sin(x) \)[/tex] is [tex]\(-1\)[/tex]. Therefore, the minimum value of [tex]\( f(x) \)[/tex] is:
[tex]\[ f_{\text{min}} = -1 + 2 = 1 \][/tex]
This occurs at [tex]\( x = 180^\circ \)[/tex] and [tex]\( x = 360^\circ \)[/tex].
4. Plot the Function Values:
- At [tex]\( x = 0^\circ \)[/tex], [tex]\( f(0) = 2 \)[/tex]
- At [tex]\( x = 90^\circ \)[/tex], [tex]\( f(90) = 3 \)[/tex]
- At [tex]\( x = 180^\circ \)[/tex], [tex]\( f(180) = 1 \)[/tex]
- At [tex]\( x = 270^\circ \)[/tex], [tex]\( f(270) = 3 \)[/tex]
- At [tex]\( x = 360^\circ \)[/tex], [tex]\( f(360) = 2 \)[/tex]
5. Sketch the Graph:
- Draw the x-axis ranging from [tex]\( 0^\circ \)[/tex] to [tex]\( 360^\circ \)[/tex].
- Draw the y-axis and mark the intercept at [tex]\((0, 2)\)[/tex].
- Plot points at [tex]\((90^\circ, 3)\)[/tex], [tex]\((180^\circ, 1)\)[/tex], [tex]\((270^\circ, 3)\)[/tex], and [tex]\((360^\circ, 2)\)[/tex].
- Sketch a smooth curve through these points to illustrate the sine wave shifted up by 2 units.
### Sketch of [tex]\( f(x) = \sin(x) + 2 \)[/tex]:
```
y
^
3 |
|
|
2 |------------------------------------------------> x
|
1 | *
|
|
0 +------------------------------------------------------
0° 90° 180° 270° 360°
```
In the graph:
- The y-intercept is clearly marked at [tex]\((0, 2)\)[/tex].
- The curve shows one full period of the sine function shifted up by 2 units, peaking at [tex]\(90^\circ\)[/tex] and [tex]\(270^\circ\)[/tex].
The sketch accurately represents [tex]\( f(x) = \sin(x) + 2 \)[/tex] over the interval [tex]\( 0^\circ \)[/tex] to [tex]\( 360^\circ \)[/tex].
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