Find the range of the following functions.
`f(x)=(5)/(3-x^(2))`

Answer

To find the range of the function \( f(x) = \frac{5}{3 - x^2} \), we will follow these steps:

### Step 1: Set the function equal to \( y \)
We start by rewriting the function in terms of \( y \):
\[ y = \frac{5}{3 - x^2} \]

### Step 2: Rearrange the equation
We can rearrange this equation to express \( x^2 \) in terms of \( y \):
\[ y(3 - x^2) = 5 \]
\[ 3y - yx^2 = 5 \]
\[ yx^2 = 3y - 5 \]
\[ x^2 = \frac{3y - 5}{y} \]

### Step 3: Analyze the expression for \( x^2 \)
Since \( x^2 \) must be non-negative (i.e., \( x^2 \geq 0 \)), we set up the inequality:
\[ \frac{3y - 5}{y} \geq 0 \]

### Step 4: Solve the inequality
To solve this inequality, we consider the two factors: \( 3y - 5 \) and \( y \).

1. **Find the critical points**:
- The numerator \( 3y - 5 = 0 \) gives \( y = \frac{5}{3} \).
- The denominator \( y = 0 \) is also a critical point.

2. **Test intervals**:
We will test the sign of \( \frac{3y - 5}{y} \) in the intervals determined by the critical points \( y = 0 \) and \( y = \frac{5}{3} \):
- For \( y < 0 \): Both \( 3y - 5 < 0 \) and \( y < 0 \) → \( \frac{3y - 5}{y} > 0 \) (positive).
- For \( 0 < y < \frac{5}{3} \): \( 3y - 5 < 0 \) and \( y > 0 \) → \( \frac{3y - 5}{y} < 0 \) (negative).
- For \( y > \frac{5}{3} \): Both \( 3y - 5 > 0 \) and \( y > 0 \) → \( \frac{3y - 5}{y} > 0 \) (positive).

### Step 5: Determine the valid intervals
From our analysis, the intervals where \( \frac{3y - 5}{y} \geq 0 \) are:
- \( y < 0 \)
- \( y > \frac{5}{3} \)

### Step 6: Write the range
Thus, the range of the function \( f(x) = \frac{5}{3 - x^2} \) is:
\[ (-\infty, 0) \cup \left(\frac{5}{3}, \infty\right) \]

### Final Answer
The range of the function \( f(x) = \frac{5}{3 - x^2} \) is:
\[ (-\infty, 0) \cup \left(\frac{5}{3}, \infty\right) \]