Let's determine the range of the function $f(x) = \sqrt{\frac{1 + x}{x}}$.

Step 1: Determine the domain of the function

For $f(x) = \sqrt{\frac{1 + x}{x}}$ to be real and defined, the expression inside the square root, $\frac{1 + x}{x}$, must be non-negative: $\frac{1 + x}{x} \geq 0$

This inequality can be solved by considering the signs of the numerator and denominator:

• $1 + x$ must have the same sign as $x$, i.e., both should be positive or both should be negative.

Case 1: $x > 0$

If $x > 0$, we require $1 + x \geq 0$, which is always true for $x > 0$. So, for positive values of $x$, the expression is defined.

Case 2: $x < 0$

If $x < 0$, we require $1 + x \leq 0$, or $x \leq -1$. Therefore, $x$ must be less than or equal to -1 for the expression to be valid.

Thus, the domain of $f(x)$ is: $x \in (-\infty, -1] \cup (0, \infty)$

Step 2: Analyze the range

Now let's determine the behavior of the function on these intervals.

Interval 1: $x > 0$

For $x > 0$, the expression $\frac{1 + x}{x}$ simplifies to: $\frac{1 + x}{x} = 1 + \frac{1}{x}$ As $x \to 0^+$, $\frac{1}{x} \to \infty$, so $\sqrt{1 + \frac{1}{x}} \to \infty$.

As $x \to \infty$, $\frac{1}{x} \to 0$, so $\sqrt{1 + \frac{1}{x}} \to \sqrt{1} = 1$.

Thus, as $x$ increases from 0 to $\infty$, the function value decreases from infinity to 1. Therefore, the range of $f(x)$ for $x > 0$ is: $(1, \infty)$

Interval 2: $x \leq -1$

For $x \leq -1$, the expression $\frac{1 + x}{x}$ simplifies to: $\frac{1 + x}{x} = 1 + \frac{1}{x}$ As $x \to -1^-$, $\frac{1}{x} \to -1$, so $\sqrt{1 + \frac{1}{x}} \to \sqrt{0} = 0$.

As $x \to -\infty$, $\frac{1}{x} \to 0$, so $\sqrt{1 + \frac{1}{x}} \to \sqrt{1} = 1$.

Thus, as $x$ decreases from -1 to $-\infty$, the function value increases from 0 to 1. Therefore, the range of $f(x)$ for $x \leq -1$ is: $[0, 1)$

Step 3: Conclusion

Combining the two intervals, the full range of the function $f(x) = \sqrt{\frac{1 + x}{x}}$ is: $[0, 1) \cup (1, \infty)$

This means the function takes all values greater than or equal to 0, except 1.

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Would you like further details on any part of this solution? Here are some related questions to explore:

1. How would the behavior of the function change if we modified the denominator to $x + 1$?
2. What happens to the range of $f(x) = \sqrt{\frac{1 + x}{x^2}}$?
3. How can we graphically visualize the function $f(x) = \sqrt{\frac{1 + x}{x}}$?
4. Can we find the inverse of $f(x) = \sqrt{\frac{1 + x}{x}}$?
5. What is the domain and range of $f(x) = \sqrt{x + 1}$?

Tip: To analyze the range of functions with square roots, always check the non-negativity of the expression inside the root first!