Let f(x) be a linear function which maps [-1, 1] onto [0, 2], then f(x) can be

Options

A. `x+1`
B. `-x-1`
C. `-x+1`
D. `x-1`

Answer

To solve the problem, we need to find a linear function \( f(x) \) that maps the interval \([-1, 1]\) onto the interval \([0, 2]\). A linear function can be expressed in the form:

\[ f(x) = mx + c \]

where \( m \) is the slope and \( c \) is the y-intercept.

### Step 1: Determine the endpoints of the function
We know that \( f(-1) \) should equal the minimum of the range, which is 0, and \( f(1) \) should equal the maximum of the range, which is 2.

\[ f(-1) = m(-1) + c = 0 \quad \text{(1)} \]
\[ f(1) = m(1) + c = 2 \quad \text{(2)} \]

### Step 2: Set up the equations
From equation (1):

\[ -c + m = 0 \implies c = m \quad \text{(3)} \]

From equation (2):

\[ m + c = 2 \quad \text{(4)} \]

### Step 3: Substitute equation (3) into equation (4)
Substituting \( c = m \) into equation (4):

\[ m + m = 2 \implies 2m = 2 \implies m = 1 \]

### Step 4: Find the value of \( c \)
Now substituting \( m = 1 \) back into equation (3):

\[ c = 1 \]

### Step 5: Write the function
Thus, the function \( f(x) \) is:

\[ f(x) = 1x + 1 = x + 1 \]

### Step 6: Verify the function
Now we need to verify that this function maps \([-1, 1]\) onto \([0, 2]\):

- For \( x = -1 \):
\[ f(-1) = -1 + 1 = 0 \]

- For \( x = 1 \):
\[ f(1) = 1 + 1 = 2 \]

Since \( f(x) \) is a linear function with a positive slope, it will cover all values between 0 and 2 as \( x \) varies from -1 to 1.

### Step 7: Check other options
We can also check the other proposed functions:

1. **Option 2**: \( f(x) = -x - 1 \)
- \( f(-1) = 0 \)
- \( f(1) = -2 \) (not valid)

2. **Option 3**: \( f(x) = -x + 1 \)
- \( f(-1) = 2 \)
- \( f(1) = 0 \) (valid, range is [0, 2])

3. **Option 4**: \( f(x) = x - 1 \)
- \( f(-1) = -2 \)
- \( f(1) = 0 \) (not valid)

### Conclusion
The valid functions that map \([-1, 1]\) onto \([0, 2]\) are:

- \( f(x) = x + 1 \)
- \( f(x) = -x + 1 \)

Thus, the final answer is that \( f(x) \) can be \( x + 1 \) or \( -x + 1 \).

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