Given f(x) = sqrt(x+1) and g(x) = 1/x, give a form for fog and gof, hence find the domain and range of each.

Given $f(x) = \sqrt{x+1}$ and $g(x) = \frac{1}{x}$.
The composite function $f \circ g (x)$ is defined as $f(g(x))$.

Substitute $\frac{1}{x}$ into $f(x)$:

To simplify the expression under the square root, find a common denominator:

The formula for $f \circ g (x)$ is $\boxed{\sqrt{\frac{1+x}{x}}}$.

For $f(g(x)) = \sqrt{\frac{1+x}{x}}$ to be defined, two conditions must be met:

1. The denominator cannot be zero: $x \neq 0$.
2. The expression under the square root must be non-negative: $\frac{1+x}{x} \ge 0$. This inequality holds when:
   • Case 1: $1+x \ge 0$ and $x > 0$. This implies $x \ge -1$ and $x > 0$, which simplifies to $x > 0$.
   • Case 2: $1+x \le 0$ and $x < 0$. This implies $x \le -1$ and $x < 0$, which simplifies to $x \le -1$. Combining these, the domain is $(-\infty, -1] \cup (0, \infty)$. The domain of $f \circ g (x)$ is $\boxed{(-\infty, -1] \cup (0, \infty)}$.

Let $y = \sqrt{\frac{1+x}{x}}$. Since $y$ is the principal square root, $y \ge 0$.
Consider the two parts of the domain:

• If $x > 0$: As $x \to 0^+$, $\frac{1+x}{x} \to \infty$, so $y \to \infty$. As $x \to \infty$, $\frac{1+x}{x} = 1 + \frac{1}{x} \to 1^+$, so $y \to 1^+$. Thus, for $x > 0$, the range is $(1, \infty)$.
• If $x \le -1$: Let $x = -z$ where $z \ge 1$. Then $y = \sqrt{\frac{1-z}{-z}} = \sqrt{\frac{z-1}{z}} = \sqrt{1 - \frac{1}{z}}$. When $x = -1$ (i.e., $z=1$), $y = \sqrt{1 - \frac{1}{1}} = \sqrt{0} = 0$. As $x \to -\infty$ (i.e., $z \to \infty$), $1 - \frac{1}{z} \to 1^-$, so $y \to 1^-$. Thus, for $x \le -1$, the range is $[0, 1)$. Combining both parts, the range of $f \circ g (x)$ is $[0, 1) \cup (1, \infty)$. The range of $f \circ g (x)$ is $\boxed{[0, 1) \cup (1, \infty)}$.

The composite function $g \circ f (x)$ is defined as $g(f(x))$.

Substitute $\sqrt{x+1}$ into $g(x)$:

The formula for $g \circ f (x)$ is $\boxed{\frac{1}{\sqrt{x+1}}}$.

For $g(f(x)) = \frac{1}{\sqrt{x+1}}$ to be defined, two conditions must be met:

1. The expression under the square root must be non-negative: $x+1 \ge 0 \implies x \ge -1$.
2. The denominator cannot be zero: $\sqrt{x+1} \neq 0 \implies x+1 \neq 0 \implies x \neq -1$. Combining these, we must have $x > -1$. The domain of $g \circ f (x)$ is $\boxed{(-1, \infty)}$.

Let $y = \frac{1}{\sqrt{x+1}}$.
Since $\sqrt{x+1}$ is always positive for $x > -1$, $y$ must be positive. So $y > 0$.
As $x \to -1^+$ (approaching from the right), $\sqrt{x+1} \to 0^+$, so $y \to \infty$.
As $x \to \infty$, $\sqrt{x+1} \to \infty$, so $y \to 0^+$.
Thus, the range of $g \circ f (x)$ is $(0, \infty)$.
The range of $g \circ f (x)$ is $\boxed{(0, \infty)}$.