There are exactly two distinct linear functions, which map [-1,1] onto [0,3]. Find the point of intersection of the two functions.

Answer

To solve the problem, we need to find the point of intersection of two distinct linear functions that map the interval \([-1, 1]\) onto \([0, 3]\).

### Step-by-Step Solution:

1. **Understanding the Functions**:
We are looking for two linear functions of the form \(f(x) = ax + b\) that map the interval \([-1, 1]\) to \([0, 3]\). This means:
- \(f(-1) = 0\)
- \(f(1) = 3\)

2. **Setting Up the Equations**:
For the first function, we can set up the equations based on the mappings:
- From \(f(-1) = 0\):
\[ -a + b = 0 \quad \text{(Equation 1)} \]
- From \(f(1) = 3\):
\[ a + b = 3 \quad \text{(Equation 2)} \]

3. **Solving the Equations**:
Now we can solve these two equations simultaneously.
- From Equation 1, we can express \(b\) in terms of \(a\):
\[ b = a \quad \text{(Substituting into Equation 2)} \]
- Substitute \(b = a\) into Equation 2:
\[ a + a = 3 \implies 2a = 3 \implies a = \frac{3}{2} \]
- Substitute \(a\) back into Equation 1 to find \(b\):
\[ b = \frac{3}{2} \]

4. **First Function**:
Thus, the first linear function is:
\[ f_1(x) = \frac{3}{2}x + \frac{3}{2} \]

5. **Finding the Second Function**:
For the second function, we can switch the mappings:
- Let \(f(-1) = 3\) and \(f(1) = 0\):
- From \(f(-1) = 3\):
\[ -a + b = 3 \quad \text{(Equation 3)} \]
- From \(f(1) = 0\):
\[ a + b = 0 \quad \text{(Equation 4)} \]

6. **Solving the Second Set of Equations**:
- From Equation 4, express \(b\) in terms of \(a\):
\[ b = -a \quad \text{(Substituting into Equation 3)} \]
- Substitute \(b = -a\) into Equation 3:
\[ -a - a = 3 \implies -2a = 3 \implies a = -\frac{3}{2} \]
- Substitute \(a\) back into Equation 4 to find \(b\):
\[ b = \frac{3}{2} \]

7. **Second Function**:
Thus, the second linear function is:
\[ f_2(x) = -\frac{3}{2}x + \frac{3}{2} \]

8. **Finding the Point of Intersection**:
To find the point of intersection of \(f_1(x)\) and \(f_2(x)\), set them equal to each other:
\[ \frac{3}{2}x + \frac{3}{2} = -\frac{3}{2}x + \frac{3}{2} \]
- Rearranging gives:
\[ \frac{3}{2}x + \frac{3}{2}x = 0 \implies 3x = 0 \implies x = 0 \]
- Substitute \(x = 0\) into either function to find \(y\):
\[ f_1(0) = \frac{3}{2}(0) + \frac{3}{2} = \frac{3}{2} \]

9. **Final Answer**:
The point of intersection of the two functions is:
\[ (0, \frac{3}{2}) \]