Answer :
To find the common solution points for the functions [tex]\( f(x) = x - 1 \)[/tex] and [tex]\( g(x) = \frac{1}{3}x^2 - \frac{1}{3} \)[/tex], we need to determine where these two functions intersect.
First, let's break down the problem:
### Step 1: Define the Functions
- [tex]\( f(x) = x - 1 \)[/tex]
- [tex]\( g(x) = \frac{1}{3}x^2 - \frac{1}{3} \)[/tex]
### Step 2: Identify Potential Points
We have been provided with several potential points of intersection:
1. [tex]\((2, 1)\)[/tex]
2. [tex]\((1, 0)\)[/tex]
3. [tex]\((2, 3)\)[/tex]
4. [tex]\((5, 1)\)[/tex]
5. [tex]\((1, 2)\)[/tex]
6. [tex]\((0, 1)\)[/tex]
### Step 3: Verify Each Point
To determine if a given point [tex]\((x, y)\)[/tex] is a solution where the graphs of [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] intersect, the y-value must be the same for both functions at that x-value. Let's check each point:
1. Point [tex]\((2, 1)\)[/tex]:
- For [tex]\( f(x) = x - 1 \)[/tex]:
[tex]\[ f(2) = 2 - 1 = 1 \][/tex]
- For [tex]\( g(x) = \frac{1}{3}x^2 - \frac{1}{3} \)[/tex]:
[tex]\[ g(2) = \frac{1}{3}(2^2) - \frac{1}{3} = \frac{4}{3} - \frac{1}{3} = \frac{3}{3} = 1 \][/tex]
- Both functions yield [tex]\( y = 1 \)[/tex]. So, [tex]\((2, 1)\)[/tex] is a common solution.
2. Point [tex]\((1, 0)\)[/tex]:
- For [tex]\( f(x) = x - 1 \)[/tex]:
[tex]\[ f(1) = 1 - 1 = 0 \][/tex]
- For [tex]\( g(x) = \frac{1}{3}x^2 - \frac{1}{3} \)[/tex]:
[tex]\[ g(1) = \frac{1}{3}(1^2) - \frac{1}{3} = \frac{1}{3} - \frac{1}{3} = 0 \][/tex]
- Both functions yield [tex]\( y = 0 \)[/tex]. So, [tex]\((1, 0)\)[/tex] is a common solution.
3. Point [tex]\((2, 3)\)[/tex]:
- For [tex]\( f(x) = x - 1 \)[/tex]:
[tex]\[ f(2) = 2 - 1 = 1 \][/tex]
- For [tex]\( g(x) = \frac{1}{3}x^2 - \frac{1}{3} \)[/tex]:
[tex]\[ g(2) = \frac{1}{3}(2^2) - \frac{1}{3} = \frac{4}{3} - \frac{1}{3} = 1 \][/tex]
- Since [tex]\( y = 1 \)[/tex] for both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex], [tex]\((2, 3)\)[/tex] does not satisfy both functions simultaneously.
4. Point [tex]\((5, 1)\)[/tex]:
- For [tex]\( f(x) = x - 1 \)[/tex]:
[tex]\[ f(5) = 5 - 1 = 4 \][/tex]
- For [tex]\( g(x) = \frac{1}{3}x^2 - \frac{1}{3} \)[/tex]:
[tex]\[ g(5) = \frac{1}{3}(5^2) - \frac{1}{3} = \frac{25}{3} - \frac{1}{3} = \frac{24}{3} = 8 \][/tex]
- Neither [tex]\( f(x) \)[/tex] nor [tex]\( g(x) \)[/tex] yield [tex]\( y = 1 \)[/tex]. Therefore, [tex]\((5, 1)\)[/tex] is not a solution.
5. Point [tex]\((1, 2)\)[/tex]:
- For [tex]\( f(x) = x - 1 \)[/tex]:
[tex]\[ f(1) = 1 - 1 = 0 \][/tex]
- For [tex]\( g(x) = \frac{1}{3}x^2 - \frac{1}{3} \)[/tex]:
[tex]\[ g(1) = \frac{1}{3}(1^2) - \frac{1}{3} = 0 \][/tex]
- Since [tex]\( y = 0 \)[/tex], [tex]\((1, 2)\)[/tex] does not satisfy both functions.
6. Point [tex]\((0, 1)\)[/tex]:
- For [tex]\( f(x) = x - 1 \)[/tex]:
[tex]\[ f(0) = 0 - 1 = -1 \][/tex]
- For [tex]\( g(x) = \frac{1}{3}x^2 - \frac{1}{3} \)[/tex]:
[tex]\[ g(0) = \frac{1}{3}(0^2) - \frac{1}{3} = -\frac{1}{3} \][/tex]
- Both functions do not yield [tex]\( y = 1 \)[/tex]. Therefore, [tex]\((0, 1)\)[/tex] is not a solution.
### Conclusion
- The solutions are the points where both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] yield the same [tex]\( y \)[/tex]-value. These points are:
- [tex]\((2, 1)\)[/tex]
- [tex]\((1, 0)\)[/tex]
So, the answer is:
- [tex]\((2, 1)\)[/tex] and [tex]\((1, 0)\)[/tex]
First, let's break down the problem:
### Step 1: Define the Functions
- [tex]\( f(x) = x - 1 \)[/tex]
- [tex]\( g(x) = \frac{1}{3}x^2 - \frac{1}{3} \)[/tex]
### Step 2: Identify Potential Points
We have been provided with several potential points of intersection:
1. [tex]\((2, 1)\)[/tex]
2. [tex]\((1, 0)\)[/tex]
3. [tex]\((2, 3)\)[/tex]
4. [tex]\((5, 1)\)[/tex]
5. [tex]\((1, 2)\)[/tex]
6. [tex]\((0, 1)\)[/tex]
### Step 3: Verify Each Point
To determine if a given point [tex]\((x, y)\)[/tex] is a solution where the graphs of [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] intersect, the y-value must be the same for both functions at that x-value. Let's check each point:
1. Point [tex]\((2, 1)\)[/tex]:
- For [tex]\( f(x) = x - 1 \)[/tex]:
[tex]\[ f(2) = 2 - 1 = 1 \][/tex]
- For [tex]\( g(x) = \frac{1}{3}x^2 - \frac{1}{3} \)[/tex]:
[tex]\[ g(2) = \frac{1}{3}(2^2) - \frac{1}{3} = \frac{4}{3} - \frac{1}{3} = \frac{3}{3} = 1 \][/tex]
- Both functions yield [tex]\( y = 1 \)[/tex]. So, [tex]\((2, 1)\)[/tex] is a common solution.
2. Point [tex]\((1, 0)\)[/tex]:
- For [tex]\( f(x) = x - 1 \)[/tex]:
[tex]\[ f(1) = 1 - 1 = 0 \][/tex]
- For [tex]\( g(x) = \frac{1}{3}x^2 - \frac{1}{3} \)[/tex]:
[tex]\[ g(1) = \frac{1}{3}(1^2) - \frac{1}{3} = \frac{1}{3} - \frac{1}{3} = 0 \][/tex]
- Both functions yield [tex]\( y = 0 \)[/tex]. So, [tex]\((1, 0)\)[/tex] is a common solution.
3. Point [tex]\((2, 3)\)[/tex]:
- For [tex]\( f(x) = x - 1 \)[/tex]:
[tex]\[ f(2) = 2 - 1 = 1 \][/tex]
- For [tex]\( g(x) = \frac{1}{3}x^2 - \frac{1}{3} \)[/tex]:
[tex]\[ g(2) = \frac{1}{3}(2^2) - \frac{1}{3} = \frac{4}{3} - \frac{1}{3} = 1 \][/tex]
- Since [tex]\( y = 1 \)[/tex] for both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex], [tex]\((2, 3)\)[/tex] does not satisfy both functions simultaneously.
4. Point [tex]\((5, 1)\)[/tex]:
- For [tex]\( f(x) = x - 1 \)[/tex]:
[tex]\[ f(5) = 5 - 1 = 4 \][/tex]
- For [tex]\( g(x) = \frac{1}{3}x^2 - \frac{1}{3} \)[/tex]:
[tex]\[ g(5) = \frac{1}{3}(5^2) - \frac{1}{3} = \frac{25}{3} - \frac{1}{3} = \frac{24}{3} = 8 \][/tex]
- Neither [tex]\( f(x) \)[/tex] nor [tex]\( g(x) \)[/tex] yield [tex]\( y = 1 \)[/tex]. Therefore, [tex]\((5, 1)\)[/tex] is not a solution.
5. Point [tex]\((1, 2)\)[/tex]:
- For [tex]\( f(x) = x - 1 \)[/tex]:
[tex]\[ f(1) = 1 - 1 = 0 \][/tex]
- For [tex]\( g(x) = \frac{1}{3}x^2 - \frac{1}{3} \)[/tex]:
[tex]\[ g(1) = \frac{1}{3}(1^2) - \frac{1}{3} = 0 \][/tex]
- Since [tex]\( y = 0 \)[/tex], [tex]\((1, 2)\)[/tex] does not satisfy both functions.
6. Point [tex]\((0, 1)\)[/tex]:
- For [tex]\( f(x) = x - 1 \)[/tex]:
[tex]\[ f(0) = 0 - 1 = -1 \][/tex]
- For [tex]\( g(x) = \frac{1}{3}x^2 - \frac{1}{3} \)[/tex]:
[tex]\[ g(0) = \frac{1}{3}(0^2) - \frac{1}{3} = -\frac{1}{3} \][/tex]
- Both functions do not yield [tex]\( y = 1 \)[/tex]. Therefore, [tex]\((0, 1)\)[/tex] is not a solution.
### Conclusion
- The solutions are the points where both [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] yield the same [tex]\( y \)[/tex]-value. These points are:
- [tex]\((2, 1)\)[/tex]
- [tex]\((1, 0)\)[/tex]
So, the answer is:
- [tex]\((2, 1)\)[/tex] and [tex]\((1, 0)\)[/tex]
