`f(x)=x^2-3x`,then the points at which `f(x)=f^`(x)` are

Options

A. `1,3`
B. `1,-3`
C. `-1,3`
D. none of these

Explanation

To solve the problem, we need to find the points at which \( f(x) = f'(x) \) for the function \( f(x) = x^2 - 3x \).

### Step-by-Step Solution:

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

2. **Differentiate the function**:
To find \( f'(x) \), we differentiate \( f(x) \) with respect to \( x \):
\[ f'(x) = \frac{d}{dx}(x^2 - 3x) = 2x - 3 \]

3. **Set up the equation**:
We need to find the points where \( f(x) = f'(x) \):
\[ x^2 - 3x = 2x - 3 \]

4. **Rearrange the equation**:
Move all terms to one side of the equation:
\[ x^2 - 3x - 2x + 3 = 0 \]
Simplifying gives:
\[ x^2 - 5x + 3 = 0 \]

5. **Use the quadratic formula**:
The quadratic formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -5 \), and \( c = 3 \). Plugging in these values:
\[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1} \]
This simplifies to:
\[ x = \frac{5 \pm \sqrt{25 - 12}}{2} \]
\[ x = \frac{5 \pm \sqrt{13}}{2} \]

6. **Final values**:
Thus, the points at which \( f(x) = f'(x) \) are:
\[ x = \frac{5 + \sqrt{13}}{2} \quad \text{and} \quad x = \frac{5 - \sqrt{13}}{2} \]

### Summary of the solution:
The points at which \( f(x) = f'(x) \) are:
\[ x = \frac{5 + \sqrt{13}}{2} \quad \text{and} \quad x = \frac{5 - \sqrt{13}}{2} \]