Answer :
Let's graph the piecewise function
[tex]\[ f(x)=\begin{cases} 2x & \text{if } x \leq 3, \\ \frac{1}{3}x^2 - 2x + 9 & \text{if } x > 3. \end{cases} \][/tex]
### Step-by-Step Solution
1. Understand the piecewise definition:
- For [tex]\( x \leq 3 \)[/tex], the function is a linear function [tex]\( f(x) = 2x \)[/tex].
- For [tex]\( x > 3 \)[/tex], the function is a quadratic function [tex]\( f(x) = \frac{1}{3}x^2 - 2x + 9 \)[/tex].
2. Determine key points and intervals:
- The boundary point is [tex]\( x = 3 \)[/tex].
- We need to evaluate [tex]\( f(x) \)[/tex] at the boundary point to ensure the function is continuous:
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ f(3) = 2 \cdot 3 = 6 \quad \text{(from } f(x) = 2x \text{)}. \][/tex]
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ f(3) = \frac{1}{3}(3)^2 - 2(3) + 9 = \frac{1}{3}(9) - 6 + 9 = 3 - 6 + 9 = 6 \quad \text{(from } f(x) = \frac{1}{3}x^2 - 2x + 9 \text{)}. \][/tex]
- Both conditions give the same value [tex]\( f(3) = 6 \)[/tex], confirming [tex]\( f(x) \)[/tex] is continuous at [tex]\( x = 3 \)[/tex].
3. Graph the linear part ([tex]\( x \leq 3 \)[/tex]):
- This is a line with slope [tex]\( 2 \)[/tex] passing through the origin.
- Compute a few key values:
[tex]\[ f(0) = 2 \cdot 0 = 0, \quad f(1) = 2 \cdot 1 = 2, \quad f(2) = 2 \cdot 2 = 4, \quad f(3) = 2 \cdot 3 = 6. \][/tex]
- Plot points: (0,0), (1,2), (2,4), (3,6).
4. Graph the quadratic part ([tex]\( x > 3 \)[/tex]):
- This is a parabola opening upwards.
- Compute a few values to help plot the curve:
[tex]\[ f(4) = \frac{1}{3}(4)^2 - 2(4) + 9 = \frac{1}{3}(16) - 8 + 9 = \frac{16}{3} - 8 + 9 = \frac{16}{3} - \frac{24}{3} + \frac{27}{3} = \frac{19}{3} \approx 6.333, \][/tex]
[tex]\[ f(5) = \frac{1}{3}(5)^2 - 2(5) + 9 = \frac{1}{3}(25) - 10 + 9 = \frac{25}{3} - 10 + 9 = \frac{25}{3} - \frac{30}{3} + \frac{27}{3} = \frac{22}{3} \approx 7.333. \][/tex]
- Plot points: (4,[tex]\(\frac{19}{3} \approx 6.333\)[/tex]), (5,[tex]\(\frac{22}{3} \approx 7.333\)[/tex]), etc.
5. Combine the parts into a single plot:
- Draw the line [tex]\( y = 2x \)[/tex] from [tex]\( x = -\infty \)[/tex] to [tex]\( x = 3 \)[/tex].
- Draw the parabola [tex]\( y = \frac{1}{3}x^2 - 2x + 9 \)[/tex] starting from [tex]\( x = 3 \)[/tex] onwards.
### Final Graph
On the graph:
- The line segment [tex]\( y = 2x \)[/tex] starts from negative infinity and goes up to the point [tex]\( (3, 6) \)[/tex].
- A sense of continuity exists at [tex]\( x = 3 \)[/tex] where the value of [tex]\( f \)[/tex] is [tex]\( 6 \)[/tex].
- The quadratic segment starts from [tex]\( (3, 6) \)[/tex] and curves upwards continuing for [tex]\( x > 3 \)[/tex].
To plot the graph:
1. Mark points on the [tex]\( y = 2x \)[/tex] line for [tex]\( x \leq 3 \)[/tex].
2. Mark points on the quadratic curve for [tex]\( x > 3 \)[/tex].
3. Clearly indicate the boundary at [tex]\( x = 3 \)[/tex] where both parts meet smoothly at [tex]\((3, 6)\)[/tex].
### Graph:
```plaintext
y
↑
|
10+
|
|
6+----------------------
|
|
|
0+----------------*------> x
-4 -3 0 3 5
Line: y = 2x → Quadratic: y = (1/3)x^2 - 2x + 9
```
[tex]\[ f(x)=\begin{cases} 2x & \text{if } x \leq 3, \\ \frac{1}{3}x^2 - 2x + 9 & \text{if } x > 3. \end{cases} \][/tex]
### Step-by-Step Solution
1. Understand the piecewise definition:
- For [tex]\( x \leq 3 \)[/tex], the function is a linear function [tex]\( f(x) = 2x \)[/tex].
- For [tex]\( x > 3 \)[/tex], the function is a quadratic function [tex]\( f(x) = \frac{1}{3}x^2 - 2x + 9 \)[/tex].
2. Determine key points and intervals:
- The boundary point is [tex]\( x = 3 \)[/tex].
- We need to evaluate [tex]\( f(x) \)[/tex] at the boundary point to ensure the function is continuous:
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ f(3) = 2 \cdot 3 = 6 \quad \text{(from } f(x) = 2x \text{)}. \][/tex]
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ f(3) = \frac{1}{3}(3)^2 - 2(3) + 9 = \frac{1}{3}(9) - 6 + 9 = 3 - 6 + 9 = 6 \quad \text{(from } f(x) = \frac{1}{3}x^2 - 2x + 9 \text{)}. \][/tex]
- Both conditions give the same value [tex]\( f(3) = 6 \)[/tex], confirming [tex]\( f(x) \)[/tex] is continuous at [tex]\( x = 3 \)[/tex].
3. Graph the linear part ([tex]\( x \leq 3 \)[/tex]):
- This is a line with slope [tex]\( 2 \)[/tex] passing through the origin.
- Compute a few key values:
[tex]\[ f(0) = 2 \cdot 0 = 0, \quad f(1) = 2 \cdot 1 = 2, \quad f(2) = 2 \cdot 2 = 4, \quad f(3) = 2 \cdot 3 = 6. \][/tex]
- Plot points: (0,0), (1,2), (2,4), (3,6).
4. Graph the quadratic part ([tex]\( x > 3 \)[/tex]):
- This is a parabola opening upwards.
- Compute a few values to help plot the curve:
[tex]\[ f(4) = \frac{1}{3}(4)^2 - 2(4) + 9 = \frac{1}{3}(16) - 8 + 9 = \frac{16}{3} - 8 + 9 = \frac{16}{3} - \frac{24}{3} + \frac{27}{3} = \frac{19}{3} \approx 6.333, \][/tex]
[tex]\[ f(5) = \frac{1}{3}(5)^2 - 2(5) + 9 = \frac{1}{3}(25) - 10 + 9 = \frac{25}{3} - 10 + 9 = \frac{25}{3} - \frac{30}{3} + \frac{27}{3} = \frac{22}{3} \approx 7.333. \][/tex]
- Plot points: (4,[tex]\(\frac{19}{3} \approx 6.333\)[/tex]), (5,[tex]\(\frac{22}{3} \approx 7.333\)[/tex]), etc.
5. Combine the parts into a single plot:
- Draw the line [tex]\( y = 2x \)[/tex] from [tex]\( x = -\infty \)[/tex] to [tex]\( x = 3 \)[/tex].
- Draw the parabola [tex]\( y = \frac{1}{3}x^2 - 2x + 9 \)[/tex] starting from [tex]\( x = 3 \)[/tex] onwards.
### Final Graph
On the graph:
- The line segment [tex]\( y = 2x \)[/tex] starts from negative infinity and goes up to the point [tex]\( (3, 6) \)[/tex].
- A sense of continuity exists at [tex]\( x = 3 \)[/tex] where the value of [tex]\( f \)[/tex] is [tex]\( 6 \)[/tex].
- The quadratic segment starts from [tex]\( (3, 6) \)[/tex] and curves upwards continuing for [tex]\( x > 3 \)[/tex].
To plot the graph:
1. Mark points on the [tex]\( y = 2x \)[/tex] line for [tex]\( x \leq 3 \)[/tex].
2. Mark points on the quadratic curve for [tex]\( x > 3 \)[/tex].
3. Clearly indicate the boundary at [tex]\( x = 3 \)[/tex] where both parts meet smoothly at [tex]\((3, 6)\)[/tex].
### Graph:
```plaintext
y
↑
|
10+
|
|
6+----------------------
|
|
|
0+----------------*------> x
-4 -3 0 3 5
Line: y = 2x → Quadratic: y = (1/3)x^2 - 2x + 9
```
