At which points on the curve `y=x/((1+x^(2)))`, the tangent is parallel to the x-axis.

Options

A. `(1, 1/2)`
B. `(pm1, 1/2)`
C. `(1/2, pm1)`
D. `(pm1, pm1/2)`

Answer

To find the points on the curve \( y = \frac{x}{1 + x^2} \) where the tangent is parallel to the x-axis, we need to determine where the slope of the curve is zero. The slope of the curve can be found by calculating the derivative \( \frac{dy}{dx} \).

### Step-by-Step Solution:

1. **Differentiate the function**:
We have \( y = \frac{x}{1 + x^2} \). To find the derivative \( \frac{dy}{dx} \), we will use the quotient rule, which states that if \( y = \frac{u}{v} \), then \( \frac{dy}{dx} = \frac{u'v - uv'}{v^2} \).

Here, \( u = x \) and \( v = 1 + x^2 \).

- Calculate \( u' \) and \( v' \):
- \( u' = 1 \)
- \( v' = 2x \)

- Apply the quotient rule:
\[ \frac{dy}{dx} = \frac{(1)(1 + x^2) - (x)(2x)}{(1 + x^2)^2} \]
Simplifying this gives:
\[ \frac{dy}{dx} = \frac{1 + x^2 - 2x^2}{(1 + x^2)^2} = \frac{1 - x^2}{(1 + x^2)^2} \]

2. **Set the derivative equal to zero**:
For the tangent to be parallel to the x-axis, we need:
\[ \frac{dy}{dx} = 0 \]
This occurs when the numerator is zero:
\[ 1 - x^2 = 0 \]
Solving for \( x \):
\[ x^2 = 1 \implies x = \pm 1 \]

3. **Find the corresponding y-values**:
Now we substitute \( x = 1 \) and \( x = -1 \) back into the original equation to find the y-coordinates.

- For \( x = 1 \):
\[ y = \frac{1}{1 + 1^2} = \frac{1}{2} \]
So one point is \( (1, \frac{1}{2}) \).

- For \( x = -1 \):
\[ y = \frac{-1}{1 + (-1)^2} = \frac{-1}{2} \]
So the other point is \( (-1, -\frac{1}{2}) \).

4. **Final points**:
The points on the curve where the tangent is parallel to the x-axis are:
\[ (1, \frac{1}{2}) \quad \text{and} \quad (-1, -\frac{1}{2}) \]

### Summary of the Solution:
The points on the curve \( y = \frac{x}{1 + x^2} \) where the tangent is parallel to the x-axis are \( (1, \frac{1}{2}) \) and \( (-1, -\frac{1}{2}) \).