Find the range of ,`f(x)=1/(4-x^2)`

Answer

To find the range of the function \( f(x) = \frac{1}{4 - x^2} \), follow these steps:

### Step 1: Express \( f(x) \) in terms of \( y \)
Given:
\[ f(x) = \frac{1}{4 - x^2} \]
Let:
\[ y = \frac{1}{4 - x^2} \]

### Step 2: Solve for \( x^2 \) in terms of \( y \)
Rearrange the equation to solve for \( x^2 \):
\[ y(4 - x^2) = 1 \]
\[ 4y - yx^2 = 1 \]
\[ yx^2 = 4y - 1 \]
\[ x^2 = \frac{4y - 1}{y} \]

### Step 3: Determine the conditions for \( x^2 \) to be non-negative
Since \( x^2 \geq 0 \), the right-hand side must be non-negative:
\[ \frac{4y - 1}{y} \geq 0 \]

### Step 4: Analyze the inequality
Consider the inequality \( \frac{4y - 1}{y} \geq 0 \):
- The numerator \( 4y - 1 \) must be non-negative.
- The denominator \( y \) must be positive or negative, but not zero.

Thus, we have two cases:
1. \( 4y - 1 \geq 0 \) and \( y > 0 \)
2. \( 4y - 1 \leq 0 \) and \( y < 0 \)

### Step 5: Solve the inequalities
1. For \( 4y - 1 \geq 0 \):
\[ 4y \geq 1 \]
\[ y \geq \frac{1}{4} \]

2. For \( 4y - 1 \leq 0 \):
\[ 4y \leq 1 \]
\[ y \leq \frac{1}{4} \]

Since \( y \neq 0 \), we have:
\[ y < 0 \]

### Step 6: Combine the intervals
The range of \( y \) is:
\[ y \in (-\infty, 0) \cup \left[\frac{1}{4}, \infty\right) \]

### Final Answer
The range of the function \( f(x) = \frac{1}{4 - x^2} \) is:
\[ (-\infty, 0) \cup \left[\frac{1}{4}, \infty\right) \]