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).
Value of b for the tangent drawn to the curve y=f(x) whose slope is greatest, is

Options

A. `(9)/(8)`
B. `(3)/(8)`
C. `(1)/(8)`
D. `(5)/(8)`

Answer

To solve the problem, we need to find the y-intercept \( b \) of the tangent to the curve \( f(x) = \frac{1}{1+x^2} \) that has the greatest slope.

### Step-by-Step Solution:

1. **Find the derivative \( f'(x) \)**:
The slope of the tangent line to the curve at any point is given by the derivative of the function. We will differentiate \( f(x) \).

\[ f'(x) = \frac{d}{dx} \left( \frac{1}{1+x^2} \right) \]

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

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

2. **Find the critical points for maximum slope**:
To find the maximum slope, we need to find the critical points of \( f'(x) \). Set \( f'(x) = 0 \):

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

Next, we check the second derivative or analyze the behavior of \( f'(x) \) around \( x = 0 \) to confirm if it is a maximum.

3. **Evaluate the slope at \( x = 0 \)**:
Substitute \( x = 0 \) into \( f'(x) \):

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

Since \( f'(x) \) changes from positive to negative around \( x = 0 \), this indicates a local maximum.

4. **Find the coordinates of the point on the curve**:
Now, we find the coordinates of the point on the curve at \( x = 0 \):

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

Thus, the point is \( (0, 1) \).

5. **Write the equation of the tangent line**:
The slope at this point is \( 0 \), so the equation of the tangent line is:

\[ y - 1 = 0 \cdot (x - 0) \implies y = 1 \]

6. **Find the y-intercept \( b \)**:
The y-intercept \( b \) of the tangent line is simply the value of \( y \) when \( x = 0 \):

\[ b = 1 \]

### Final Answer:
The value of \( b \) for the tangent drawn to the curve \( y = f(x) \) whose slope is greatest is \( \boxed{1} \).

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