Let `f(x)=(1)/(1+x^(2)),` let m be the slope, a be the x-intercept and b be they y-intercept of a tangent to y=f(x).
Absicca of the point of contact of the tangent for which m is greatest, is

Options

A. `(1)/sqrt(3)`
B. 1
C. -1
D. `-(1)/sqrt(3)`

Answer

To solve the problem step by step, we will find the x-coordinate of the point of contact of the tangent to the curve \( f(x) = \frac{1}{1+x^2} \) where the slope \( m \) is greatest.

### Step 1: Differentiate the function
We start by finding the derivative \( f'(x) \) to determine the slope of the tangent line at any point \( x \).

\[ f(x) = \frac{1}{1+x^2} \]

Using the quotient rule, where \( u = 1 \) and \( v = 1 + x^2 \):

\[ f'(x) = \frac{u'v - uv'}{v^2} \]

Calculating \( u' \) and \( v' \):

- \( u' = 0 \)
- \( v' = 2x \)

Now substituting into the quotient rule:

\[ f'(x) = \frac{0 \cdot (1 + x^2) - 1 \cdot (2x)}{(1 + x^2)^2} = \frac{-2x}{(1 + x^2)^2} \]

### Step 2: Set the derivative to find critical points
To find the maximum slope, we need to find the critical points by setting \( f'(x) = 0 \):

\[ \frac{-2x}{(1 + x^2)^2} = 0 \]

This implies:

\[ -2x = 0 \implies x = 0 \]

### Step 3: Analyze the behavior of the slope
Next, we need to determine if this critical point is a maximum. We can check the second derivative or analyze the first derivative around this point.

However, since we are looking for the maximum slope, we can also consider the endpoints or the behavior of the function as \( x \) approaches positive and negative infinity.

### Step 4: Find the maximum slope
To find the maximum value of the slope, we can evaluate \( f'(x) \) at \( x = 0 \):

\[ f'(0) = \frac{-2(0)}{(1 + 0^2)^2} = 0 \]

Now we need to check the limits as \( x \) approaches \( \pm \infty \):

\[ \lim_{x \to \infty} f'(x) = \lim_{x \to \infty} \frac{-2x}{(1+x^2)^2} = 0 \]

\[ \lim_{x \to -\infty} f'(x) = \lim_{x \to -\infty} \frac{-2x}{(1+x^2)^2} = 0 \]

### Step 5: Find the maximum value of the slope
To find the maximum slope, we can also check the second derivative or use the first derivative test. However, we can also consider the nature of the function \( f'(x) \):

The slope \( f'(x) \) is negative for \( x > 0 \) and positive for \( x < 0 \). Hence, the maximum slope occurs at \( x = -\frac{1}{\sqrt{3}} \) and \( x = \frac{1}{\sqrt{3}} \).

### Conclusion
Thus, the x-coordinate of the point of contact of the tangent for which the slope \( m \) is greatest is:

\[ x = -\frac{1}{\sqrt{3}} \]