The function `f (x)= ( x)/( 4 +x^2)` decreases in the interval

Options

A. `(1 ,2)`
B. `(0,e)`
C. `(-oo ,2)` U `( 2,oo)`
D. `(1//e,2e)`

Answer

To determine the interval in which the function \( f(x) = \frac{x}{4 + x^2} \) is decreasing, we need to follow these steps:

### Step 1: Differentiate the function
We will differentiate \( f(x) \) using the quotient rule. The quotient rule states that if you have a function in the form \( \frac{u}{v} \), then its derivative is given by:

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

Here, \( u = x \) and \( v = 4 + x^2 \).

- \( u' = 1 \) (the derivative of \( x \))
- \( v' = 2x \) (the derivative of \( 4 + x^2 \))

Now, applying the quotient rule:

\[ f'(x) = \frac{(1)(4 + x^2) - (x)(2x)}{(4 + x^2)^2} \]

### Step 2: Simplify the derivative
Now we simplify the expression:

\[ f'(x) = \frac{4 + x^2 - 2x^2}{(4 + x^2)^2} = \frac{4 - x^2}{(4 + x^2)^2} \]

### Step 3: Determine where the derivative is less than zero
To find where the function is decreasing, we need to solve the inequality:

\[ f'(x) < 0 \]

This occurs when the numerator \( 4 - x^2 < 0 \) because the denominator \( (4 + x^2)^2 \) is always positive.

### Step 4: Solve the inequality
Solving \( 4 - x^2 < 0 \):

\[ 4 < x^2 \]

Taking square roots gives:

\[ |x| > 2 \]

This means:

\[ x < -2 \quad \text{or} \quad x > 2 \]

### Step 5: Write the interval
Thus, the function \( f(x) \) is decreasing in the intervals:

\[ (-\infty, -2) \cup (2, \infty) \]

### Conclusion
The intervals in which the function \( f(x) = \frac{x}{4 + x^2} \) is decreasing are:

\[ (-\infty, -2) \cup (2, \infty) \]