The range of the function, `f (x) = (1+ x ^(2))/(x ^(2))` is

Options

A. `(0,1)`
B. `[0,1]`
C. `(1,oo)`
D. `[1,oo)`

Answer

To find the range of the function \( f(x) = \frac{1 + x^2}{x^2} \), we can simplify and analyze the function step by step.

### Step 1: Rewrite the function
We can rewrite the function as:
\[ f(x) = \frac{1}{x^2} + 1 \]

### Step 2: Analyze the term \( \frac{1}{x^2} \)
The term \( \frac{1}{x^2} \) is defined for all \( x \neq 0 \). Since \( x^2 \) is always positive for all \( x \neq 0 \), \( \frac{1}{x^2} \) will always be positive. As \( x \) approaches 0 (from either side), \( \frac{1}{x^2} \) approaches infinity. As \( |x| \) increases, \( \frac{1}{x^2} \) approaches 0.

### Step 3: Determine the minimum value of \( f(x) \)
From our analysis:
- As \( x \) approaches 0, \( f(x) \) approaches infinity.
- As \( |x| \) increases, \( f(x) \) approaches \( 1 \) (since \( \frac{1}{x^2} \) approaches 0).

Thus, the minimum value of \( f(x) \) occurs as \( |x| \) becomes very large, and \( f(x) \) approaches \( 1 \).

### Step 4: Determine the range
Since \( f(x) \) approaches \( 1 \) but never actually reaches it (because \( \frac{1}{x^2} \) is always positive), the function can take any value greater than \( 1 \) and can go up to infinity.

Thus, the range of the function is:
\[ (1, \infty) \]

### Final Answer
The range of the function \( f(x) = \frac{1 + x^2}{x^2} \) is \( (1, \infty) \).

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