Let `f(x)=(e^(x))/(1+x^(2)) and g(x)=f'(x)` , then

Options

A. g(x) has two local maxima and two local minima points
B. g(x) has exactly one local maxima and one local minima point
C. x = 1 is a point of local maxima for g(x)
D. There is a point of local maxima for g(x) in the interval (-1,0)

Explanation

To solve the problem, we need to analyze the function \( f(x) = \frac{e^x}{1 + x^2} \) and its derivative \( g(x) = f'(x) \) to find the maximum and minimum values.

### Step 1: Find the derivative of \( f(x) \)

Using the quotient rule for differentiation, we have:

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

Simplifying this, we get:

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

### Step 2: Set the derivative \( g(x) = f'(x) \) to zero

To find critical points, we set the numerator equal to zero:

\[ 1 - 2x + x^2 = 0 \]

This is a quadratic equation. We can solve it using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 - 4}}{2} = \frac{2 \pm 0}{2} = 1 \]

### Step 3: Analyze the critical point

The only critical point we found is \( x = 1 \). To determine whether this point is a maximum or minimum, we can analyze the sign of \( f'(x) \) around this point.

### Step 4: Use the first derivative test

We can check the sign of \( f'(x) \) for values less than and greater than 1:

- For \( x < 1 \), say \( x = 0 \):
\[ f'(0) = \frac{e^0(1 - 0)}{(1 + 0)^2} = 1 > 0 \quad (\text{increasing}) \]

- For \( x > 1 \), say \( x = 2 \):
\[ f'(2) = \frac{e^2(1 - 4 + 4)}{(1 + 4)^2} = \frac{e^2(1)}{25} > 0 \quad (\text{increasing}) \]

Since \( f'(x) \) changes from positive to negative at \( x = 1 \), we conclude that \( x = 1 \) is a local maximum.

### Step 5: Determine the behavior as \( x \to \pm \infty \)

As \( x \to \infty \):
\[ f(x) \to 0 \quad \text{(since \( e^x \) grows faster than \( 1 + x^2 \))} \]

As \( x \to -\infty \):
\[ f(x) \to 0 \quad \text{(since \( e^x \to 0 \))} \]

### Conclusion

The function \( f(x) \) has a local maximum at \( x = 1 \) and approaches 0 as \( x \to \pm \infty \). Therefore, the maximum value occurs at \( x = 1 \).