Show that there lies a point on the curve `f(x)=x^(2)-4x +3` between (1,0) and (3,0) where tangent drawn is parallel to x-axis. Find its coordinates.

Answer

To solve the problem, we need to show that there exists a point on the curve \( f(x) = x^2 - 4x + 3 \) between the points (1, 0) and (3, 0) where the tangent drawn is parallel to the x-axis. A tangent is parallel to the x-axis when its slope is zero.

### Step-by-Step Solution:

1. **Find the derivative of the function**:
The function is given by:
\[ f(x) = x^2 - 4x + 3 \]
We need to find the derivative \( f'(x) \) to determine the slope of the tangent.
\[ f'(x) = \frac{d}{dx}(x^2 - 4x + 3) = 2x - 4 \]

2. **Set the derivative equal to zero**:
To find the points where the tangent is parallel to the x-axis, we set the derivative equal to zero:
\[ 2x - 4 = 0 \]
Solving for \( x \):
\[ 2x = 4 \implies x = 2 \]

3. **Find the corresponding y-coordinate**:
Now, we substitute \( x = 2 \) back into the original function to find the y-coordinate:
\[ f(2) = 2^2 - 4(2) + 3 = 4 - 8 + 3 = -1 \]
Thus, the coordinates of the point are \( (2, -1) \).

4. **Verify that the point lies between (1, 0) and (3, 0)**:
The point \( (2, -1) \) lies between \( (1, 0) \) and \( (3, 0) \) on the x-axis, as \( 1 < 2 < 3 \).

### Conclusion:
We have shown that there exists a point on the curve \( f(x) = x^2 - 4x + 3 \) between (1, 0) and (3, 0) where the tangent is parallel to the x-axis. The coordinates of this point are \( (2, -1) \).