Answer :
Sure, let's graph the piecewise-defined function:
[tex]\[ f(x)=\left\{\begin{array}{ll} -x+3, & x<2 \\ 3, & 2 \leq x<4 \\ 4-2 x, & x \geq 4 \end{array}\right. \][/tex]
### Step 1: Understanding Each Piece
1. Piece 1: [tex]\( f(x) = -x + 3 \)[/tex] for [tex]\( x < 2 \)[/tex]
This is a linear function with a negative slope. The y-intercept is at [tex]\( (0, 3) \)[/tex].
2. Piece 2: [tex]\( f(x) = 3 \)[/tex] for [tex]\( 2 \leq x < 4 \)[/tex]
This is a constant function, so for all values of [tex]\( x \)[/tex] in the interval [tex]\( [2, 4) \)[/tex], [tex]\( f(x) = 3 \)[/tex].
3. Piece 3: [tex]\( f(x) = 4 - 2x \)[/tex] for [tex]\( x \geq 4 \)[/tex]
This is another linear function with a negative slope, and it starts at [tex]\( x = 4 \)[/tex].
### Step 2: Plotting Each Piece
1. For [tex]\( x < 2 \)[/tex], [tex]\( f(x) = -x + 3 \)[/tex]
- Start by plotting the line [tex]\( y = -x + 3 \)[/tex] for values of [tex]\( x \)[/tex] less than 2.
- At [tex]\( x = 2 \)[/tex], [tex]\( f(x) \)[/tex] would be [tex]\( -2 + 3 = 1 \)[/tex].
- This endpoint ([tex]\( x = 2 \)[/tex]) will be an open circle because [tex]\( x = 2 \)[/tex] does not include this point.
2. For [tex]\( 2 \leq x < 4 \)[/tex], [tex]\( f(x) = 3 \)[/tex]
- At [tex]\( x = 2 \)[/tex], [tex]\( f(x) = 3 \)[/tex] (since [tex]\( x = 2 \)[/tex] is included in this interval).
- This endpoint ([tex]\( x = 2 \)[/tex]) will be a closed circle at [tex]\( (2, 3) \)[/tex].
- This is a horizontal line segment from [tex]\( x = 2 \)[/tex] to [tex]\( x = 4 \)[/tex], not including 4.
- At [tex]\( x = 4 \)[/tex], [tex]\( f(x) = 3 \)[/tex]. This endpoint will be an open circle because [tex]\( x = 4 \)[/tex] is not included in this piece.
3. For [tex]\( x \geq 4 \)[/tex], [tex]\( f(x) = 4 - 2x \)[/tex]
- Start by plotting the line [tex]\( y = 4 - 2x \)[/tex] for values of [tex]\( x \)[/tex] greater than or equal to 4.
- At [tex]\( x = 4 \)[/tex], [tex]\( f(x) = 4 - 2(4) = 4 - 8 = -4 \)[/tex]. This endpoint ([tex]\( x = 4 \)[/tex]) will be a closed circle.
### Step 3: Combining Everything
1. Piece 1: [tex]\( f(x) = -x + 3 \)[/tex] for [tex]\( x < 2 \)[/tex]
- Plot points like [tex]\( (0, 3) \)[/tex] and [tex]\( (1, 2) \)[/tex].
- Open circle at [tex]\( (2, 1) \)[/tex].
2. Piece 2: [tex]\( f(x) = 3 \)[/tex] for [tex]\( 2 \leq x < 4 \)[/tex]
- Horizontal line from [tex]\( (2, 3) \)[/tex] to just before [tex]\( (4, 3) \)[/tex].
- Closed circle at [tex]\( (2, 3) \)[/tex].
- Open circle at [tex]\( (4, 3) \)[/tex].
3. Piece 3: [tex]\( f(x) = 4 - 2x \)[/tex] for [tex]\( x \geq 4 \)[/tex]
- Plot points like [tex]\( (4, -4) \)[/tex] and [tex]\( (5, -6) \)[/tex].
- Closed circle at [tex]\( (4, -4) \)[/tex].
By carefully plotting these pieces and marking the endpoints appropriately, you can correctly graph the piecewise function [tex]\( f(x) \)[/tex].
[tex]\[ f(x)=\left\{\begin{array}{ll} -x+3, & x<2 \\ 3, & 2 \leq x<4 \\ 4-2 x, & x \geq 4 \end{array}\right. \][/tex]
### Step 1: Understanding Each Piece
1. Piece 1: [tex]\( f(x) = -x + 3 \)[/tex] for [tex]\( x < 2 \)[/tex]
This is a linear function with a negative slope. The y-intercept is at [tex]\( (0, 3) \)[/tex].
2. Piece 2: [tex]\( f(x) = 3 \)[/tex] for [tex]\( 2 \leq x < 4 \)[/tex]
This is a constant function, so for all values of [tex]\( x \)[/tex] in the interval [tex]\( [2, 4) \)[/tex], [tex]\( f(x) = 3 \)[/tex].
3. Piece 3: [tex]\( f(x) = 4 - 2x \)[/tex] for [tex]\( x \geq 4 \)[/tex]
This is another linear function with a negative slope, and it starts at [tex]\( x = 4 \)[/tex].
### Step 2: Plotting Each Piece
1. For [tex]\( x < 2 \)[/tex], [tex]\( f(x) = -x + 3 \)[/tex]
- Start by plotting the line [tex]\( y = -x + 3 \)[/tex] for values of [tex]\( x \)[/tex] less than 2.
- At [tex]\( x = 2 \)[/tex], [tex]\( f(x) \)[/tex] would be [tex]\( -2 + 3 = 1 \)[/tex].
- This endpoint ([tex]\( x = 2 \)[/tex]) will be an open circle because [tex]\( x = 2 \)[/tex] does not include this point.
2. For [tex]\( 2 \leq x < 4 \)[/tex], [tex]\( f(x) = 3 \)[/tex]
- At [tex]\( x = 2 \)[/tex], [tex]\( f(x) = 3 \)[/tex] (since [tex]\( x = 2 \)[/tex] is included in this interval).
- This endpoint ([tex]\( x = 2 \)[/tex]) will be a closed circle at [tex]\( (2, 3) \)[/tex].
- This is a horizontal line segment from [tex]\( x = 2 \)[/tex] to [tex]\( x = 4 \)[/tex], not including 4.
- At [tex]\( x = 4 \)[/tex], [tex]\( f(x) = 3 \)[/tex]. This endpoint will be an open circle because [tex]\( x = 4 \)[/tex] is not included in this piece.
3. For [tex]\( x \geq 4 \)[/tex], [tex]\( f(x) = 4 - 2x \)[/tex]
- Start by plotting the line [tex]\( y = 4 - 2x \)[/tex] for values of [tex]\( x \)[/tex] greater than or equal to 4.
- At [tex]\( x = 4 \)[/tex], [tex]\( f(x) = 4 - 2(4) = 4 - 8 = -4 \)[/tex]. This endpoint ([tex]\( x = 4 \)[/tex]) will be a closed circle.
### Step 3: Combining Everything
1. Piece 1: [tex]\( f(x) = -x + 3 \)[/tex] for [tex]\( x < 2 \)[/tex]
- Plot points like [tex]\( (0, 3) \)[/tex] and [tex]\( (1, 2) \)[/tex].
- Open circle at [tex]\( (2, 1) \)[/tex].
2. Piece 2: [tex]\( f(x) = 3 \)[/tex] for [tex]\( 2 \leq x < 4 \)[/tex]
- Horizontal line from [tex]\( (2, 3) \)[/tex] to just before [tex]\( (4, 3) \)[/tex].
- Closed circle at [tex]\( (2, 3) \)[/tex].
- Open circle at [tex]\( (4, 3) \)[/tex].
3. Piece 3: [tex]\( f(x) = 4 - 2x \)[/tex] for [tex]\( x \geq 4 \)[/tex]
- Plot points like [tex]\( (4, -4) \)[/tex] and [tex]\( (5, -6) \)[/tex].
- Closed circle at [tex]\( (4, -4) \)[/tex].
By carefully plotting these pieces and marking the endpoints appropriately, you can correctly graph the piecewise function [tex]\( f(x) \)[/tex].
