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Sagot :
To determine the graph of the piecewise function [tex]\( f(x) \)[/tex], we divide the process into two parts according to the definition of [tex]\( f(x) \)[/tex].
### Part 1: [tex]\( f(x) = \frac{1}{2} x - 4 \)[/tex] for [tex]\(-4 \leq x \leq 4 \)[/tex]
1. Equation: [tex]\( y = \frac{1}{2} x - 4 \)[/tex]
2. Input Range: [tex]\(-4 \leq x \leq 4 \)[/tex]
3. Calculate few key points:
- When [tex]\( x = -4 \)[/tex]:
[tex]\[ y = \frac{1}{2}(-4) - 4 = -2 - 4 = -6 \][/tex]
- When [tex]\( x = 0 \)[/tex]:
[tex]\[ y = \frac{1}{2}(0) - 4 = -4 \][/tex]
- When [tex]\( x = 4 \)[/tex]:
[tex]\[ y = \frac{1}{2}(4) - 4 = 2 - 4 = -2 \][/tex]
So, the graph for this part within the domain will be a line passing through points [tex]\((-4, -6)\)[/tex], [tex]\((0, -4)\)[/tex], and [tex]\((4, -2)\)[/tex].
### Part 2: [tex]\( f(x) = 2x - 7 \)[/tex] for [tex]\(4 < x \leq 6\)[/tex]
1. Equation: [tex]\( y = 2x - 7 \)[/tex]
2. Input Range: [tex]\(4 < x \leq 6\)[/tex]
3. Calculate few key points:
- When [tex]\( x = 4 \)[/tex] (continuing for consistency, though [tex]\( x \neq 4 \)[/tex]):
[tex]\[ y = 2(4) - 7 = 8 - 7 = 1 \][/tex]
- When [tex]\( x = 5 \)[/tex]:
[tex]\[ y = 2(5) - 7 = 10 - 7 = 3 \][/tex]
- When [tex]\( x = 6 \)[/tex]:
[tex]\[ y = 2(6) - 7 = 12 - 7 = 5 \][/tex]
So, the graph for this part within the domain will be a line passing through points [tex]\((4, 1)\)[/tex], [tex]\((5, 3)\)[/tex], and [tex]\((6, 5)\)[/tex].
### Combining Both Parts
- From [tex]\(-4\)[/tex] to [tex]\(\ 4\)[/tex], the function is [tex]\( y = \frac{1}{2}x - 4 \)[/tex], a straight line starting at point [tex]\((-4, -6)\)[/tex] and ending at point [tex]\((4, -2)\)[/tex].
- From [tex]\(\ 4\)[/tex] to [tex]\(\ 6\)[/tex], the function is [tex]\( y = 2x - 7 \)[/tex], a straight line starting just after [tex]\((4, 1)\)[/tex] and ending at point [tex]\((6, 5)\)[/tex].
- There will be a discontinuity at [tex]\( x = 4 \)[/tex] since [tex]\( f(4) = -2 \)[/tex] from the first part, and jumps to [tex]\( y = 1 \)[/tex] in the second part just after 4.
Now, to visualize the graph:
1. Draw the first part: A straight line passing through points [tex]\((-4, -6)\)[/tex] to [tex]\((4, -2)\)[/tex]. This line appears to decrease in the negative y-direction until [tex]\( x \)[/tex] reaches 4.
2. Discontinuity at [tex]\( x = 4 \)[/tex]:
- Open circle at [tex]\((4, -2)\)[/tex], indicating that the first part of the function doesn't include this point.
3. Draw the second part: A straight line passing through points just after [tex]\((4, 1)\)[/tex], marking a clear discontinuity with an open circle at [tex]\( (4, 1) \)[/tex], to [tex]\((6, 5)\)[/tex].
Conclusively, the graph of [tex]\( f(x) \)[/tex] will encompass these characteristics, showing a continuous line for [tex]\(-4 \leq x \leq 4\)[/tex] transforming into another continuous line for [tex]\( 4 < x \leq 6\)[/tex], separated by a jump discontinuity between the ranges at x = 4.
Thus, the correct answer should show:
- A straight line decreasing from [tex]\((4, -2)\)[/tex] to [tex]\((-4, -6)\)[/tex]
- A straight line increasing from an open circle at [tex]\((4, 1)\)[/tex] to [tex]\((6, 5)\)[/tex]
The option that visually matches this description and includes these points as calculated is the correct answer.
### Part 1: [tex]\( f(x) = \frac{1}{2} x - 4 \)[/tex] for [tex]\(-4 \leq x \leq 4 \)[/tex]
1. Equation: [tex]\( y = \frac{1}{2} x - 4 \)[/tex]
2. Input Range: [tex]\(-4 \leq x \leq 4 \)[/tex]
3. Calculate few key points:
- When [tex]\( x = -4 \)[/tex]:
[tex]\[ y = \frac{1}{2}(-4) - 4 = -2 - 4 = -6 \][/tex]
- When [tex]\( x = 0 \)[/tex]:
[tex]\[ y = \frac{1}{2}(0) - 4 = -4 \][/tex]
- When [tex]\( x = 4 \)[/tex]:
[tex]\[ y = \frac{1}{2}(4) - 4 = 2 - 4 = -2 \][/tex]
So, the graph for this part within the domain will be a line passing through points [tex]\((-4, -6)\)[/tex], [tex]\((0, -4)\)[/tex], and [tex]\((4, -2)\)[/tex].
### Part 2: [tex]\( f(x) = 2x - 7 \)[/tex] for [tex]\(4 < x \leq 6\)[/tex]
1. Equation: [tex]\( y = 2x - 7 \)[/tex]
2. Input Range: [tex]\(4 < x \leq 6\)[/tex]
3. Calculate few key points:
- When [tex]\( x = 4 \)[/tex] (continuing for consistency, though [tex]\( x \neq 4 \)[/tex]):
[tex]\[ y = 2(4) - 7 = 8 - 7 = 1 \][/tex]
- When [tex]\( x = 5 \)[/tex]:
[tex]\[ y = 2(5) - 7 = 10 - 7 = 3 \][/tex]
- When [tex]\( x = 6 \)[/tex]:
[tex]\[ y = 2(6) - 7 = 12 - 7 = 5 \][/tex]
So, the graph for this part within the domain will be a line passing through points [tex]\((4, 1)\)[/tex], [tex]\((5, 3)\)[/tex], and [tex]\((6, 5)\)[/tex].
### Combining Both Parts
- From [tex]\(-4\)[/tex] to [tex]\(\ 4\)[/tex], the function is [tex]\( y = \frac{1}{2}x - 4 \)[/tex], a straight line starting at point [tex]\((-4, -6)\)[/tex] and ending at point [tex]\((4, -2)\)[/tex].
- From [tex]\(\ 4\)[/tex] to [tex]\(\ 6\)[/tex], the function is [tex]\( y = 2x - 7 \)[/tex], a straight line starting just after [tex]\((4, 1)\)[/tex] and ending at point [tex]\((6, 5)\)[/tex].
- There will be a discontinuity at [tex]\( x = 4 \)[/tex] since [tex]\( f(4) = -2 \)[/tex] from the first part, and jumps to [tex]\( y = 1 \)[/tex] in the second part just after 4.
Now, to visualize the graph:
1. Draw the first part: A straight line passing through points [tex]\((-4, -6)\)[/tex] to [tex]\((4, -2)\)[/tex]. This line appears to decrease in the negative y-direction until [tex]\( x \)[/tex] reaches 4.
2. Discontinuity at [tex]\( x = 4 \)[/tex]:
- Open circle at [tex]\((4, -2)\)[/tex], indicating that the first part of the function doesn't include this point.
3. Draw the second part: A straight line passing through points just after [tex]\((4, 1)\)[/tex], marking a clear discontinuity with an open circle at [tex]\( (4, 1) \)[/tex], to [tex]\((6, 5)\)[/tex].
Conclusively, the graph of [tex]\( f(x) \)[/tex] will encompass these characteristics, showing a continuous line for [tex]\(-4 \leq x \leq 4\)[/tex] transforming into another continuous line for [tex]\( 4 < x \leq 6\)[/tex], separated by a jump discontinuity between the ranges at x = 4.
Thus, the correct answer should show:
- A straight line decreasing from [tex]\((4, -2)\)[/tex] to [tex]\((-4, -6)\)[/tex]
- A straight line increasing from an open circle at [tex]\((4, 1)\)[/tex] to [tex]\((6, 5)\)[/tex]
The option that visually matches this description and includes these points as calculated is the correct answer.
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