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Sagot :
To graph a piecewise-defined function, it's important to carefully analyze each piece of the function separately and understand the conditions under which each piece is defined. Let's go through the given function step-by-step:
[tex]\[ f(x) = \begin{cases} -x + 3 & \text{for } x < 2 \\ 3 & \text{for } 2 \leq x < 4 \\ 4 - 2x & \text{for } x \geq 4 \end{cases} \][/tex]
### Step-by-Step Instructions:
1. Identify the intervals and corresponding function expressions:
- For [tex]\( x < 2 \)[/tex]: [tex]\( f(x) = -x + 3 \)[/tex]
- For [tex]\( 2 \leq x < 4 \)[/tex]: [tex]\( f(x) = 3 \)[/tex]
- For [tex]\( x \geq 4 \)[/tex]: [tex]\( f(x) = 4 - 2x \)[/tex]
2. Graph each piece on the specified intervals:
- For [tex]\( x < 2 \)[/tex]:
- The function is [tex]\( f(x) = -x + 3 \)[/tex].
- This is a linear function with a slope of [tex]\(-1\)[/tex] and a y-intercept of [tex]\(3\)[/tex].
- Graph this line for values of [tex]\( x \)[/tex] less than [tex]\( 2 \)[/tex].
- The endpoint at [tex]\( x = 2 \)[/tex] will be an open circle because [tex]\( x \)[/tex] does not include [tex]\( 2 \)[/tex] (i.e., [tex]\( x < 2 \)[/tex]).
- For [tex]\( 2 \leq x < 4 \)[/tex]:
- The function is [tex]\( f(x) = 3 \)[/tex].
- This is a constant function where [tex]\( f(x) \)[/tex] stays at [tex]\( 3 \)[/tex] for all [tex]\( x \)[/tex] in the interval [tex]\( [2, 4) \)[/tex].
- Draw a horizontal line at [tex]\( y = 3 \)[/tex] from [tex]\( x = 2 \)[/tex] to [tex]\( x < 4 \)[/tex].
- The endpoint at [tex]\( x = 2 \)[/tex] will be a closed circle because [tex]\( x \)[/tex] includes [tex]\( 2 \)[/tex] (i.e., [tex]\( 2 \leq x \)[/tex]).
- The endpoint at [tex]\( x = 4 \)[/tex] will be an open circle because [tex]\( x \)[/tex] does not include [tex]\( 4 \)[/tex] (i.e., [tex]\( x < 4 \)[/tex]).
- For [tex]\( x \geq 4 \)[/tex]:
- The function is [tex]\( f(x) = 4 - 2x \)[/tex].
- This is a linear function with a slope of [tex]\(-2\)[/tex] and a y-intercept of [tex]\( 4 \)[/tex].
- Graph this line for values of [tex]\( x \)[/tex] greater than or equal to [tex]\( 4 \)[/tex].
- The endpoint at [tex]\( x = 4 \)[/tex] will be a closed circle because [tex]\( x \)[/tex] includes [tex]\( 4 \)[/tex] (i.e., [tex]\( x \geq 4 \)[/tex]).
### Plotting Each Piece:
1. For [tex]\( x < 2 \)[/tex]:
- Start at a point less than 2, for example, [tex]\( x = 1 \)[/tex]:
- If [tex]\( x = 1 \)[/tex], then [tex]\( f(1) = -1 + 3 = 2 \)[/tex].
- Continue plotting points towards [tex]\( x = 2 \)[/tex]:
- If [tex]\( x = 0 \)[/tex], then [tex]\( f(0) = -0 + 3 = 3 \)[/tex].
- If [tex]\( x = -1 \)[/tex], then [tex]\( f(-1) = 1 + 3 = 4 \)[/tex].
- The endpoint at [tex]\( x = 2 \)[/tex] is an open circle:
- Calculate [tex]\( f(2) = -2 + 3 = 1 \)[/tex], but mark (2, 1) with an open circle.
2. For [tex]\( 2 \leq x < 4 \)[/tex]:
- Draw a horizontal line at [tex]\( y = 3 \)[/tex]:
- The line starts with a closed circle at [tex]\( (2, 3) \)[/tex] and ends with an open circle at [tex]\( (4, 3) \)[/tex].
3. For [tex]\( x \geq 4 \)[/tex]:
- Calculate at [tex]\( x = 4 \)[/tex]:
- [tex]\( f(4) = 4 - 2 \times 4 = 4 - 8 = -4 \)[/tex]
- Continue plotting points greater than 4:
- If [tex]\( x = 5 \)[/tex], then [tex]\( f(5) = 4 - 2 \times 5 = 4 - 10 = -6 \)[/tex].
- The endpoint at [tex]\( x = 4 \)[/tex] is a closed circle at (4, -4).
Now, combine all these pieces in a single graph:
- Plot the points and graph the function [tex]\( f(x) = -x + 3 \)[/tex] up to, but not including, [tex]\( x = 2 \)[/tex], marking an open circle at (2,1).
- Draw a horizontal line at [tex]\( y = 3 \)[/tex] for [tex]\( x \)[/tex] from 2 to just before 4. Place a closed circle at (2,3) and an open circle at (4,3).
- Plot the points and graph the function [tex]\( f(x) = 4 - 2x \)[/tex] starting from [tex]\( x = 4 \)[/tex] and onwards, marking a closed circle at (4, -4).
By carefully assembling these pieces, you will have an accurate graph of the given piecewise function.
[tex]\[ f(x) = \begin{cases} -x + 3 & \text{for } x < 2 \\ 3 & \text{for } 2 \leq x < 4 \\ 4 - 2x & \text{for } x \geq 4 \end{cases} \][/tex]
### Step-by-Step Instructions:
1. Identify the intervals and corresponding function expressions:
- For [tex]\( x < 2 \)[/tex]: [tex]\( f(x) = -x + 3 \)[/tex]
- For [tex]\( 2 \leq x < 4 \)[/tex]: [tex]\( f(x) = 3 \)[/tex]
- For [tex]\( x \geq 4 \)[/tex]: [tex]\( f(x) = 4 - 2x \)[/tex]
2. Graph each piece on the specified intervals:
- For [tex]\( x < 2 \)[/tex]:
- The function is [tex]\( f(x) = -x + 3 \)[/tex].
- This is a linear function with a slope of [tex]\(-1\)[/tex] and a y-intercept of [tex]\(3\)[/tex].
- Graph this line for values of [tex]\( x \)[/tex] less than [tex]\( 2 \)[/tex].
- The endpoint at [tex]\( x = 2 \)[/tex] will be an open circle because [tex]\( x \)[/tex] does not include [tex]\( 2 \)[/tex] (i.e., [tex]\( x < 2 \)[/tex]).
- For [tex]\( 2 \leq x < 4 \)[/tex]:
- The function is [tex]\( f(x) = 3 \)[/tex].
- This is a constant function where [tex]\( f(x) \)[/tex] stays at [tex]\( 3 \)[/tex] for all [tex]\( x \)[/tex] in the interval [tex]\( [2, 4) \)[/tex].
- Draw a horizontal line at [tex]\( y = 3 \)[/tex] from [tex]\( x = 2 \)[/tex] to [tex]\( x < 4 \)[/tex].
- The endpoint at [tex]\( x = 2 \)[/tex] will be a closed circle because [tex]\( x \)[/tex] includes [tex]\( 2 \)[/tex] (i.e., [tex]\( 2 \leq x \)[/tex]).
- The endpoint at [tex]\( x = 4 \)[/tex] will be an open circle because [tex]\( x \)[/tex] does not include [tex]\( 4 \)[/tex] (i.e., [tex]\( x < 4 \)[/tex]).
- For [tex]\( x \geq 4 \)[/tex]:
- The function is [tex]\( f(x) = 4 - 2x \)[/tex].
- This is a linear function with a slope of [tex]\(-2\)[/tex] and a y-intercept of [tex]\( 4 \)[/tex].
- Graph this line for values of [tex]\( x \)[/tex] greater than or equal to [tex]\( 4 \)[/tex].
- The endpoint at [tex]\( x = 4 \)[/tex] will be a closed circle because [tex]\( x \)[/tex] includes [tex]\( 4 \)[/tex] (i.e., [tex]\( x \geq 4 \)[/tex]).
### Plotting Each Piece:
1. For [tex]\( x < 2 \)[/tex]:
- Start at a point less than 2, for example, [tex]\( x = 1 \)[/tex]:
- If [tex]\( x = 1 \)[/tex], then [tex]\( f(1) = -1 + 3 = 2 \)[/tex].
- Continue plotting points towards [tex]\( x = 2 \)[/tex]:
- If [tex]\( x = 0 \)[/tex], then [tex]\( f(0) = -0 + 3 = 3 \)[/tex].
- If [tex]\( x = -1 \)[/tex], then [tex]\( f(-1) = 1 + 3 = 4 \)[/tex].
- The endpoint at [tex]\( x = 2 \)[/tex] is an open circle:
- Calculate [tex]\( f(2) = -2 + 3 = 1 \)[/tex], but mark (2, 1) with an open circle.
2. For [tex]\( 2 \leq x < 4 \)[/tex]:
- Draw a horizontal line at [tex]\( y = 3 \)[/tex]:
- The line starts with a closed circle at [tex]\( (2, 3) \)[/tex] and ends with an open circle at [tex]\( (4, 3) \)[/tex].
3. For [tex]\( x \geq 4 \)[/tex]:
- Calculate at [tex]\( x = 4 \)[/tex]:
- [tex]\( f(4) = 4 - 2 \times 4 = 4 - 8 = -4 \)[/tex]
- Continue plotting points greater than 4:
- If [tex]\( x = 5 \)[/tex], then [tex]\( f(5) = 4 - 2 \times 5 = 4 - 10 = -6 \)[/tex].
- The endpoint at [tex]\( x = 4 \)[/tex] is a closed circle at (4, -4).
Now, combine all these pieces in a single graph:
- Plot the points and graph the function [tex]\( f(x) = -x + 3 \)[/tex] up to, but not including, [tex]\( x = 2 \)[/tex], marking an open circle at (2,1).
- Draw a horizontal line at [tex]\( y = 3 \)[/tex] for [tex]\( x \)[/tex] from 2 to just before 4. Place a closed circle at (2,3) and an open circle at (4,3).
- Plot the points and graph the function [tex]\( f(x) = 4 - 2x \)[/tex] starting from [tex]\( x = 4 \)[/tex] and onwards, marking a closed circle at (4, -4).
By carefully assembling these pieces, you will have an accurate graph of the given piecewise function.
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