If `f(x)=x^2 -3x +1` find `x in R` such that `f(2x)=f(x).`

Answer

To solve the problem, we need to find the values of \( x \) in \( \mathbb{R} \) such that \( f(2x) = f(x) \) where \( f(x) = x^2 - 3x + 1 \).

### Step-by-Step Solution:

1. **Define the function**:
\[ f(x) = x^2 - 3x + 1 \]

2. **Calculate \( f(2x) \)**:
Substitute \( 2x \) into the function:
\[ f(2x) = (2x)^2 - 3(2x) + 1 \]
Simplifying this:
\[ f(2x) = 4x^2 - 6x + 1 \]

3. **Set up the equation**:
We need to set \( f(2x) \) equal to \( f(x) \):
\[ 4x^2 - 6x + 1 = x^2 - 3x + 1 \]

4. **Simplify the equation**:
Subtract \( x^2 - 3x + 1 \) from both sides:
\[ 4x^2 - 6x + 1 - (x^2 - 3x + 1) = 0 \]
This simplifies to:
\[ 4x^2 - 6x + 1 - x^2 + 3x - 1 = 0 \]
Combine like terms:
\[ (4x^2 - x^2) + (-6x + 3x) + (1 - 1) = 0 \]
Which gives:
\[ 3x^2 - 3x = 0 \]

5. **Factor the equation**:
Factor out the common term:
\[ 3x(x - 1) = 0 \]

6. **Find the solutions**:
Set each factor to zero:
\[ 3x = 0 \quad \text{or} \quad x - 1 = 0 \]
This gives:
\[ x = 0 \quad \text{or} \quad x = 1 \]

7. **Conclusion**:
The values of \( x \) that satisfy \( f(2x) = f(x) \) are:
\[ x = 0 \quad \text{and} \quad x = 1 \]

### Final Answer:
The values of \( x \) in \( \mathbb{R} \) such that \( f(2x) = f(x) \) are \( x = 0 \) and \( x = 1 \).

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